Dual Nature of Radiation and Matter

de brogle wavelength is given by

$$\begin{gathered} \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} \\ \mathrm{here}, \mathrm{wavelength} \mathrm{is} \mathrm{same} \mathrm{for} \mathrm{electron} \mathrm{and} \mathrm{proton}, \\ \mathrm{so} \mathrm{the} \mathrm{momentum} \mathrm{is} \mathrm{also} \mathrm{same} \mathrm{for} \mathrm{eletron} \mathrm{and} \mathrm{proton}. \end{gathered}$$

The energy of a single photon is given by the formula \(E=h\nu\) ,where \(h\) is Planck's constant. The value of Planck's constant is

\(h=6.626\times10^{-34}J\cdot s\).

\( E=(6.626\times10^{-34}J\cdot s)(5.0\times10^{14}Hz) \)

\( E=3.313\times10^{-19}J \)

Step 3: Calculate the number of photons emitted per second

The number of photons emitted per second, N, is the total power divided by the energy of a single photon.

\( N=\frac{P}{E} \)

\( N=\frac{3.31\times10^{-3}W}{3.313\times10^{-19}J} \)

\(N\approx9.99\times10^{15}\text{ photons/s} \)

The energy of a single photon is given by the formula \(E=h\nu\) ,where \(h\) is Planck's constant. The value of Planck's constant is

\(h=6.626\times10^{-34}J\cdot s\).

\( E=(6.626\times10^{-34}J\cdot s)(5.0\times10^{14}Hz) \)

\( E=3.313\times10^{-19}J \)

Step 3: Calculate the number of photons emitted per second

The number of photons emitted per second, N, is the total power divided by the energy of a single photon.

\( N=\frac{P}{E} \)

\( N=\frac{3.31\times10^{-3}W}{3.313\times10^{-19}J} \)

\(N\approx9.99\times10^{15}\text{ photons/s} \)

Wavelengths associated with an electron ( \(\lambda_e\) ), proton ( \(\lambda_p\) ), and deuteron ( \(\lambda_d\) ), all moving with the same speed.

Using de Broglie's equation: \(\boldsymbol{\lambda}=\frac{\boldsymbol{h}}{\boldsymbol{m} v}\)
- Since all particles have the same speed (v), their wavelengths are inversely proportional to their masses.
- Electron has the smallest mass, so it has the longest wavelength.
- Deuteron has the highest mass, so it has the shortest wavelength.

Mass relation: \(m_e<m_p<m_d\)
thus,
wavelength relation: \(\lambda_e>\lambda_p>\lambda_d\)

Momentum is given by

$$\begin{gathered} \mathrm{p}= \frac{\mathrm{E}}{\mathrm{c}} \\ \mathrm{So}, \\ \mathrm{E} = \mathrm{p} \mathrm{c} \end{gathered}$$

$$\begin{gathered} K·E=e{V}_s \\ =2eV \\ K·E·=E-\varnothing \\ 2=E-2.14 \\ E=4.14eV \\ \frac{hc}{\lambda }=E \\ \frac{1242}{\lambda }=4.14 \end{gathered}$$

$\lambda =300nm$

\begin{aligned}
& \text{Given: photon energies } h\nu_A=3.3\ \mathrm{eV},\; h\nu_B=11.3\ \mathrm{eV},\; \text{work function } \phi=2.3\ \mathrm{eV}. \\[4pt]
& \text{Maximum kinetic energy of emitted electrons: } K_{\max}=h\nu-\phi. \\[4pt]
& K_{A}=3.3-2.3=1.0\ \mathrm{eV}, \qquad K_{B}=11.3-2.3=9.0\ \mathrm{eV}. \\[6pt]
& \text{Since } K=\tfrac{1}{2}mv^2 \Rightarrow v\propto\sqrt{K}, \\[4pt]
& \therefore\; \dfrac{v_A}{v_B}=\sqrt{\dfrac{K_A}{K_B}}=\sqrt{\dfrac{1.0}{9.0}}=\dfrac{1}{3}. \\[8pt]
& \boxed{\dfrac{v_A}{v_B}=\dfrac{1}{3}}
\end{aligned}

$$\begin{gathered} \mathrm{In} \mathrm{relativistic} \mathrm{case} \\ \mathrm{m}=\frac{{\mathrm{m}}_{\mathrm{o}}}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}} \\ {\mathrm{p}}^2 = \frac{{\mathrm{m}}_{\mathrm{o}}^2 {\mathrm{v}}^2}{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}=\frac{{\mathrm{m}}_{\mathrm{o}}^2 {\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}{\mathrm{c}}^2}{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)} \\ {\mathrm{p}}^2 = \frac{{\mathrm{m}}_{\mathrm{o}}^2\left(\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2} -1+1\right) {\mathrm{c}}^2}{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}={\mathrm{m}}_{\mathrm{o}}^2{\mathrm{c}}^2 + \frac{{\mathrm{m}}_{\mathrm{o}}^2 {\mathrm{c}}^2}{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)} \\ {\mathrm{p}}^2{\mathrm{c}}^2 = {\mathrm{m}}_{\mathrm{o}}^2{\mathrm{c}}^2 + \left(\frac{{\mathrm{m}}_{\mathrm{o}}}{\sqrt{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}}\right){\mathrm{c}}^4 \\ {\mathrm{p}}^2{\mathrm{c}}^2 = -{\left({\mathrm{m}}_{\mathrm{o}}{\mathrm{c}}^2\right)}^2 + {\left({\mathrm{mc}}^2\right)}^2 \\ {\left({\mathrm{mc}}^2\right)}^2 ={\mathrm{p}}^2{\mathrm{c}}^2 + {\left({\mathrm{m}}_{\mathrm{o}}{\mathrm{c}}^2\right)}^2 \\ {\mathrm{E}}^2 = {\mathrm{p}}^2{\mathrm{c}}^2 + \left({\mathrm{m}}_{\mathrm{o}}^2{\mathrm{c}}^4\right) \\ \mathrm{E} = \sqrt{{\mathrm{p}}^2{\mathrm{c}}^2 + \left({\mathrm{m}}_{\mathrm{o}}^2{\mathrm{c}}^4\right)} \end{gathered}$$

The energy of a single photon is given by the formula \(E=h\nu\) ,where \(h\) is Planck's constant. The value of Planck's constant is

\(h=6.626\times10^{-34}J\cdot s\).

\( E=(6.626\times10^{-34}J\cdot s)(5.0\times10^{14}Hz) \)

\( E=3.313\times10^{-19}J \)

Step 3: Calculate the number of photons emitted per second

The number of photons emitted per second, N, is the total power divided by the energy of a single photon.

\( N=\frac{P}{E} \)

\( N=\frac{3.31\times10^{-3}W}{3.313\times10^{-19}J} \)

\(N\approx9.99\times10^{15}\text{ photons/s} \)

$$\begin{gathered} K.E_{max} = h\left(\nu -{\nu }_0\right) \\ \nu <{\nu }_0 , No electrons emit \\ K.E_{max} - depends on frequency \\ and stopping potential \end{gathered}$$

Photoelectric current depends on intensity of incident radiation.

$$\begin{gathered} K.E_{max} = h\left(\nu -{\nu }_0\right) \\ \nu <{\nu }_0 , No electrons emit \\ K.E_{max} - depends on frequency \\ and stopping potential \end{gathered}$$

Photoelectric current depends on intensity of incident radiation.

\begin{aligned}
& \text{Given: photon energies } h\nu_A=3.3\ \mathrm{eV},\; h\nu_B=11.3\ \mathrm{eV},\; \text{work function } \phi=2.3\ \mathrm{eV}. \\[4pt]
& \text{Maximum kinetic energy of emitted electrons: } K_{\max}=h\nu-\phi. \\[4pt]
& K_{A}=3.3-2.3=1.0\ \mathrm{eV}, \qquad K_{B}=11.3-2.3=9.0\ \mathrm{eV}. \\[6pt]
& \text{Since } K=\tfrac{1}{2}mv^2 \Rightarrow v\propto\sqrt{K}, \\[4pt]
& \therefore\; \dfrac{v_A}{v_B}=\sqrt{\dfrac{K_A}{K_B}}=\sqrt{\dfrac{1.0}{9.0}}=\dfrac{1}{3}. \\[8pt]
& \boxed{\dfrac{v_A}{v_B}=\dfrac{1}{3}}
\end{aligned}

de brogle wavelength is given by

$$\begin{gathered} \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} \\ \mathrm{here}, \mathrm{wavelength} \mathrm{is} \mathrm{same} \mathrm{for} \mathrm{electron} \mathrm{and} \mathrm{proton}, \\ \mathrm{so} \mathrm{the} \mathrm{momentum} \mathrm{is} \mathrm{also} \mathrm{same} \mathrm{for} \mathrm{eletron} \mathrm{and} \mathrm{proton}. \end{gathered}$$

The energy of light is related to its wavelength by:

$$E = \frac{hc}{\lambda}$$​

For a wavelength $\lambda$, the total energy is:

$$E_{\lambda} = \phi + K_{\text{max}}$$​

where:

• $\phi = 1 \, \text{eV}$(work function),
• $K_{\text{max}} = 2 \, \text{eV}$ (maximum kinetic energy).

Thus:

$$E_{\lambda} = 1 + 2 = 3 \, \text{eV}$$

For a wavelength $\frac{\lambda}{2}$, the energy is doubled (since energy is inversely proportional to wavelength):

$$E_{\lambda/2} = 2 \cdot E_{\lambda} = 2 \cdot 3 = 6 \, \text{eV}$$

The maximum kinetic energy is:

$$K_{\text{max}} = E_{\lambda/2} - \phi$$

Substitute $E_{\lambda/2} = 6 \, \text{eV}$ and $\phi = 1 \, \text{eV}$:

$$K_{\text{max}} = 6 - 1 = 5 \, \text{eV}$$

The maximum kinetic energy is:

$$\boxed{5 \, \text{eV}}$$​

Wavelengths associated with an electron ( \(\lambda_e\) ), proton ( \(\lambda_p\) ), and deuteron ( \(\lambda_d\) ), all moving with the same speed.

Using de Broglie's equation: \(\boldsymbol{\lambda}=\frac{\boldsymbol{h}}{\boldsymbol{m} v}\)
- Since all particles have the same speed (v), their wavelengths are inversely proportional to their masses.
- Electron has the smallest mass, so it has the longest wavelength.
- Deuteron has the highest mass, so it has the shortest wavelength.

Mass relation: \(m_e<m_p<m_d\)
thus,
wavelength relation: \(\lambda_e>\lambda_p>\lambda_d\)

$$\begin{gathered} {\mathrm{E}}_{\mathrm{A}} = 3.3 \mathrm{eV}, {\mathrm{E}}_{\mathrm{B}} = 11.3 \mathrm{eV}, \mathrm{W} = 2.3 \mathrm{eV} \\ \mathrm{By} \mathrm{Einstein}\text{'}\mathrm{s} \mathrm{equation} \\ \mathrm{E} = \mathrm{W} + {\mathrm{KE}}_{\max } \\ \mathrm{So}, \\ \frac{{\mathrm{KE}}_{\mathrm{A}}}{{\mathrm{KE}}_{\mathrm{B}}} = \frac{3.3 - 2.3}{11.3 - 2.3}=\frac{1}{9} \\ \frac{{\displaystyle \frac{1}{2}}{\mathrm{mv}}_{\mathrm{A}}^2}{\frac{1}{2}{\mathrm{mv}}_{\mathrm{B}}^2}=\frac{1}{9} \\ \frac{{\displaystyle {\mathrm{v}}_{\mathrm{A}}}}{{\mathrm{v}}_{\mathrm{B}}}=\frac{1}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{Energy} \mathrm{associated} \mathrm{to} \mathrm{\lambda } = 550 \mathrm{nm} \mathrm{is} \\ \mathrm{E}\hspace{0.17em}= \frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }} \\ \mathrm{E} = \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{550\times {10}^{-9}}=3.61\times {10}^{-19} \mathrm{J} \\ \mathrm{E}\hspace{0.17em}= \frac{3.61\times {10}^{-19} }{1.6\times {10}^{-19} }=2.25 \mathrm{eV} \end{gathered}$$

This energy is more than the work function of cesium and less than the work fucntion of lithium. So photoelectric emission is possible for cesium only.

The energy of the incident photon ($E$) is related to the work function and the stopping potential by:

$$E = \phi + eV$$E

Here:

• $\phi = 3.4 \, \text{eV}$
• $eV = 1.6 \, \text{eV}$
• The total energy of the incident photon:

$$E = 3.4 + 1.6 = 5.0 \, \text{eV}$$

The energy of the photon is also given by:

$$E = \frac{h c}{\lambda}$$​

Rearranging for $\lambda$:

$$\lambda = \frac{h c}{E}$$​

Substitute the given values:

$$\lambda = \frac{12400}{5} = 2480 \, \text{\AA}$$

Convert to nanometers (since $1 \, \text{\AA} = 0.1 \, \text{nm}$1A˚=0.1nm):

$$\lambda = 248 \, \text{nm}$$λ=248nm

Correct option (b) 248nm

\begin{aligned}
& \text{Given: photon energies } h\nu_A=3.3\ \mathrm{eV},\; h\nu_B=11.3\ \mathrm{eV},\; \text{work function } \phi=2.3\ \mathrm{eV}. \\[4pt]
& \text{Maximum kinetic energy of emitted electrons: } K_{\max}=h\nu-\phi. \\[4pt]
& K_{A}=3.3-2.3=1.0\ \mathrm{eV}, \qquad K_{B}=11.3-2.3=9.0\ \mathrm{eV}. \\[6pt]
& \text{Since } K=\tfrac{1}{2}mv^2 \Rightarrow v\propto\sqrt{K}, \\[4pt]
& \therefore\; \dfrac{v_A}{v_B}=\sqrt{\dfrac{K_A}{K_B}}=\sqrt{\dfrac{1.0}{9.0}}=\dfrac{1}{3}. \\[8pt]
& \boxed{\dfrac{v_A}{v_B}=\dfrac{1}{3}}
\end{aligned}

For photoemission,
\[
hf_i=\phi+K_{\max}, \qquad \phi=\frac{hc}{\lambda_0}, \qquad f_i=\frac{c}{\lambda_i}.
\]
Hence the kinetic energy of emitted electrons is
\[
K_{\max}=hc\!\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right).
\]

Using de Broglie and non-relativistic relations,
\[
K_{\max}=\frac{p^2}{2m}, \qquad \lambda_e=\frac{h}{p}.
\]
Therefore
\[
p=\sqrt{2mK_{\max}}
=\sqrt{2mhc\!\left(\frac{1}{\lambda_i}-\frac{1}{\lambda_0}\right)},
\]
and
\[
\boxed{\;\lambda_e
=\frac{h}{p}
=\sqrt{\frac{h}{2mc\!\left(\dfrac{1}{\lambda_i}-\dfrac{1}{\lambda_0}\right)}}\;}
\]

Energy of incoming photon

$=hv=W+K_{\max }$

The stopping potential $V_s$ is related to the maximum kinetic energy by:

$e{V}_s=K_{\max }$

Correct answer is (a).

Both (A) and (R) are true but (R) is not the correct explanation of (A)

$$\begin{gathered} \mathrm{In} \mathrm{relativistic} \mathrm{case} \\ \mathrm{m}=\frac{{\mathrm{m}}_{\mathrm{o}}}{\sqrt{1-{\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}}} \\ {\mathrm{p}}^2 = \frac{{\mathrm{m}}_{\mathrm{o}}^2 {\mathrm{v}}^2}{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}=\frac{{\mathrm{m}}_{\mathrm{o}}^2 {\displaystyle \frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}}{\mathrm{c}}^2}{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)} \\ {\mathrm{p}}^2 = \frac{{\mathrm{m}}_{\mathrm{o}}^2\left(\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2} -1+1\right) {\mathrm{c}}^2}{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}={\mathrm{m}}_{\mathrm{o}}^2{\mathrm{c}}^2 + \frac{{\mathrm{m}}_{\mathrm{o}}^2 {\mathrm{c}}^2}{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)} \\ {\mathrm{p}}^2{\mathrm{c}}^2 = {\mathrm{m}}_{\mathrm{o}}^2{\mathrm{c}}^2 + \left(\frac{{\mathrm{m}}_{\mathrm{o}}}{\sqrt{\left(1-\frac{{\mathrm{v}}^2}{{\mathrm{c}}^2}\right)}}\right){\mathrm{c}}^4 \\ {\mathrm{p}}^2{\mathrm{c}}^2 = -{\left({\mathrm{m}}_{\mathrm{o}}{\mathrm{c}}^2\right)}^2 + {\left({\mathrm{mc}}^2\right)}^2 \\ {\left({\mathrm{mc}}^2\right)}^2 ={\mathrm{p}}^2{\mathrm{c}}^2 + {\left({\mathrm{m}}_{\mathrm{o}}{\mathrm{c}}^2\right)}^2 \\ {\mathrm{E}}^2 = {\mathrm{p}}^2{\mathrm{c}}^2 + \left({\mathrm{m}}_{\mathrm{o}}^2{\mathrm{c}}^4\right) \\ \mathrm{E} = \sqrt{{\mathrm{p}}^2{\mathrm{c}}^2 + \left({\mathrm{m}}_{\mathrm{o}}^2{\mathrm{c}}^4\right)} \end{gathered}$$

The electron's velocity changes over time due to the electric field, therefore,

$\overrightarrow{v}={\overrightarrow{v}}_0+\left(\frac{-e{E}_{0}t}{m}\right)\hat{k}$

The new wavelength is:

$$\begin{gathered} {\lambda }^{\text{'}} =\frac{h}{mv} \\ {\lambda }^{\text{'}} =\frac{h}{m\sqrt{v_0^2+{\left(\frac{e{E}_{0}t}{m}\right)}^2}} \\ {\lambda }^{\text{'}} =\frac{{\lambda }_o}{\sqrt{1+{\left(\frac{e{E}_{0}t}{m{v}_o}\right)}^2}} where, {\lambda }_o = \frac{h}{m{v}_o} \\ {\lambda }^{\text{'}} =\frac{{\lambda }_o}{\sqrt{1+\left(\frac{e^2{E}_o^2{t}^2}{m^2{v}_o^2}\right)}} \end{gathered}$$

$$\begin{gathered} {\mathrm{E}}_{\mathrm{A}} = 3.3 \mathrm{eV}, {\mathrm{E}}_{\mathrm{B}} = 11.3 \mathrm{eV}, \mathrm{W} = 2.3 \mathrm{eV} \\ \mathrm{By} \mathrm{Einstein}\text{'}\mathrm{s} \mathrm{equation} \\ \mathrm{E} = \mathrm{W} + {\mathrm{KE}}_{\max } \\ \mathrm{So}, \\ \frac{{\mathrm{KE}}_{\mathrm{A}}}{{\mathrm{KE}}_{\mathrm{B}}} = \frac{3.3 - 2.3}{11.3 - 2.3}=\frac{1}{9} \\ \frac{{\displaystyle \frac{1}{2}}{\mathrm{mv}}_{\mathrm{A}}^2}{\frac{1}{2}{\mathrm{mv}}_{\mathrm{B}}^2}=\frac{1}{9} \\ \frac{{\displaystyle {\mathrm{v}}_{\mathrm{A}}}}{{\mathrm{v}}_{\mathrm{B}}}=\frac{1}{3} \end{gathered}$$

$$\begin{gathered} \mathrm{Energy} \mathrm{associated} \mathrm{to} \mathrm{\lambda } = 550 \mathrm{nm} \mathrm{is} \\ \mathrm{E}\hspace{0.17em}= \frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }} \\ \mathrm{E} = \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{550\times {10}^{-9}}=3.61\times {10}^{-19} \mathrm{J} \\ \mathrm{E}\hspace{0.17em}= \frac{3.61\times {10}^{-19} }{1.6\times {10}^{-19} }=2.25 \mathrm{eV} \end{gathered}$$

This energy is more than the work function of cesium and less than the work fucntion of lithium. So photoelectric emission is possible for cesium only.

$$\begin{gathered} \mathrm{Energy} \mathrm{associated} \mathrm{to} \mathrm{\lambda } = 550 \mathrm{nm} \mathrm{is} \\ \mathrm{E}\hspace{0.17em}= \frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }} \\ \mathrm{E} = \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{550\times {10}^{-9}}=3.61\times {10}^{-19} \mathrm{J} \\ \mathrm{E}\hspace{0.17em}= \frac{3.61\times {10}^{-19} }{1.6\times {10}^{-19} }=2.25 \mathrm{eV} \end{gathered}$$

This energy is more than the work function of cesium and less than the work fucntion of lithium. So photoelectric emission is possible for cesium only.

The photoelectric equation is given by:

\(K E_{\max }=h \nu-\phi\)

\(K E_{\max }=e V_0\)

where \(e\) is the charge of the electron.

\(e V_0=h \nu-\phi\)

\(V_0=\frac{h}{e} \nu-\frac{\phi}{e}\)

Comparing with the equation of a straight line \(y=m x+c\), the slope \(m\) is:

\(m=\frac{h}{e}\)

Thus, the charge of an electron \(e\) is:

\(e=\frac{h}{m}\)

Hence, the correct answer is \(\frac{h}{m}\).

Wavelengths associated with an electron ( \(\lambda_e\) ), proton ( \(\lambda_p\) ), and deuteron ( \(\lambda_d\) ), all moving with the same speed.

Using de Broglie's equation: \(\boldsymbol{\lambda}=\frac{\boldsymbol{h}}{\boldsymbol{m} v}\)

- Since all particles have the same speed (v), their wavelengths are inversely proportional to their masses.

- Electron has the smallest mass, so it has the longest wavelength.

- Deuteron has the highest mass, so it has the shortest wavelength.

Mass relation: \(m_e<m_p<m_d\)

thus,

wavelength relation: \(\lambda_e>\lambda_p>\lambda_d\)

$$\begin{gathered} K·E=e{V}_s \\ =2eV \\ K·E·=E-\varnothing \\ 2=E-2.14 \\ E=4.14eV \\ \frac{hc}{\lambda }=E \\ \frac{1242}{\lambda }=4.14 \end{gathered}$$

$\lambda =300nm$

$\frac{{\left(\mathrm{K\alpha }\right)}_{\mathrm{z}=46}}{{\mathrm{\lambda }}_{\mathrm{cutoff}}}=2$ .....(1)

$\frac{{\left(\mathrm{K\alpha }\right)}_{\mathrm{z}=41}}{{\mathrm{\lambda }}_{\mathrm{cutoff}}}=\mathrm{x}$ .....(2)

Dividing both equation

$\frac{1/{(46-1)}^2}{1/{(41-1)}^2}=\frac{2}{\mathrm{x}}$

x = 2.53

The photoelectric equation is given by:

\(K E_{\max }=h \nu-\phi\)

\(K E_{\max }=e V_0\)

where \(e\) is the charge of the electron.

\(e V_0=h \nu-\phi\)

\(V_0=\frac{h}{e} \nu-\frac{\phi}{e}\)

Comparing with the equation of a straight line \(y=m x+c\), the slope \(m\) is:

\(m=\frac{h}{e}\)

Thus, the charge of an electron \(e\) is:

\(e=\frac{h}{m}\)

Hence, the correct answer is \(\frac{h}{m}\).

$$\begin{gathered} \lambda =\frac{h}{mv} \\ \lambda =\frac{6.63\times {10}^{-34}}{\left({10}^{-30}kg\right)\times \left(2.21\times {10}^{6}m/s\right)} \\ \lambda =\frac{6.63\times {10}^{-34}}{2.21\times {10}^{-24}}\approx 3.0\times {10}^{-10}m \end{gathered}$$
Hence particle will behave as x -ray.

$$\begin{gathered} K.E_{max} = h\left(\nu -{\nu }_0\right) \\ \nu <{\nu }_0 , No electrons emit \\ K.E_{max} - depends on frequency \\ and stopping potential \end{gathered}$$

Photoelectric current depends on intensity of incident radiation.

\begin{aligned}
& \text{Given: photon energies } h\nu_A=3.3\ \mathrm{eV},\; h\nu_B=11.3\ \mathrm{eV},\; \text{work function } \phi=2.3\ \mathrm{eV}. \\[4pt]
& \text{Maximum kinetic energy of emitted electrons: } K_{\max}=h\nu-\phi. \\[4pt]
& K_{A}=3.3-2.3=1.0\ \mathrm{eV}, \qquad K_{B}=11.3-2.3=9.0\ \mathrm{eV}. \\[6pt]
& \text{Since } K=\tfrac{1}{2}mv^2 \Rightarrow v\propto\sqrt{K}, \\[4pt]
& \therefore\; \dfrac{v_A}{v_B}=\sqrt{\dfrac{K_A}{K_B}}=\sqrt{\dfrac{1.0}{9.0}}=\dfrac{1}{3}. \\[8pt]
& \boxed{\dfrac{v_A}{v_B}=\dfrac{1}{3}}
\end{aligned}

• Broglie wavelength λ = h / √(2mqV). For same V, λ ∝ 1/√(mq).
• Alpha particle (m_α, q_α = 2e) vs Electron (me, qe = e). Since m_α >> m_e and q_α > q_e, the product m_αq_α is much larger than m_eq_e.
• λ_e will be larger than λ_α.

(NEW NCERT 12^th Page No. 284, 285, 286)

de brogle wavelength is given by

$$\begin{gathered} \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{p}} \\ \mathrm{here}, \mathrm{wavelength} \mathrm{is} \mathrm{same} \mathrm{for} \mathrm{electron} \mathrm{and} \mathrm{proton}, \\ \mathrm{so} \mathrm{the} \mathrm{momentum} \mathrm{is} \mathrm{also} \mathrm{same} \mathrm{for} \mathrm{eletron} \mathrm{and} \mathrm{proton}. \end{gathered}$$

$$\begin{gathered} K.E_{max} = h\left(\nu -{\nu }_0\right) \\ \nu <{\nu }_0 , No electrons emit \\ K.E_{max} - depends on frequency \\ and stopping potential \end{gathered}$$

Photoelectric current depends on intensity of incident radiation.

Both (A) and (R) are true but (R) is not the correct explanation of (A)

Wavelengths associated with an electron ( \(\lambda_e\) ), proton ( \(\lambda_p\) ), and deuteron ( \(\lambda_d\) ), all moving with the same speed.

Using de Broglie's equation: \(\boldsymbol{\lambda}=\frac{\boldsymbol{h}}{\boldsymbol{m} v}\)

- Since all particles have the same speed (v), their wavelengths are inversely proportional to their masses.

- Electron has the smallest mass, so it has the longest wavelength.

- Deuteron has the highest mass, so it has the shortest wavelength.

Mass relation: \(m_e<m_p<m_d\)

thus,

wavelength relation: \(\lambda_e>\lambda_p>\lambda_d\)

$$\begin{gathered} {\mathrm{E}}_{\mathrm{A}} = 3.3 \mathrm{eV}, {\mathrm{E}}_{\mathrm{B}} = 11.3 \mathrm{eV}, \mathrm{W} = 2.3 \mathrm{eV} \\ \mathrm{By} \mathrm{Einstein}\text{'}\mathrm{s} \mathrm{equation} \\ \mathrm{E} = \mathrm{W} + {\mathrm{KE}}_{\max } \\ \mathrm{So}, \\ \frac{{\mathrm{KE}}_{\mathrm{A}}}{{\mathrm{KE}}_{\mathrm{B}}} = \frac{3.3 - 2.3}{11.3 - 2.3}=\frac{1}{9} \\ \frac{{\displaystyle \frac{1}{2}}{\mathrm{mv}}_{\mathrm{A}}^2}{\frac{1}{2}{\mathrm{mv}}_{\mathrm{B}}^2}=\frac{1}{9} \\ \frac{{\displaystyle {\mathrm{v}}_{\mathrm{A}}}}{{\mathrm{v}}_{\mathrm{B}}}=\frac{1}{3} \end{gathered}$$

The photoelectric equation is given by:

\(K E_{\max }=h \nu-\phi\)

\(K E_{\max }=e V_0\)

where \(e\) is the charge of the electron.

\(e V_0=h \nu-\phi\)

\(V_0=\frac{h}{e} \nu-\frac{\phi}{e}\)

Comparing with the equation of a straight line \(y=m x+c\), the slope \(m\) is:

\(m=\frac{h}{e}\)

Thus, the charge of an electron \(e\) is:

\(e=\frac{h}{m}\)

Hence, the correct answer is \(\frac{h}{m}\).

$$\begin{gathered} E=\frac{hc}{\lambda } \\ E=\frac{\left(6.626\times {10}^{-34}\right)\left(3.0\times {10}^8\right)}{550\times {10}^{-9}} \\ E=\frac{1.9878\times {10}^{-25}}{550\times {10}^{-9}} \\ E=3.61\times {10}^{-19}J=2.26eV \end{gathered}$$

Cesium (Cs): Work function = 1.9 eV
Since 2.26 eV (incident photon energy) > 1.9 eV , photoelectric emission is possible.

Lithium (Li): Work function = 2.5 eV
Since 2.26 eV (incident photon energy) < 2.5 eV, photoelectric emission is not possible.

$\frac{1}{2}{m}_p{v}^2=\frac{hc}{\lambda }=E$

$$\begin{gathered} {\lambda }_p=\frac{h}{\sqrt{2m_P\left(KE\right)}} \\ =\frac{h}{\sqrt{2m_P\left(\frac{hc}{\lambda }\right)}}=\sqrt{\frac{h\lambda }{2m_{P}c}} \end{gathered}$$

Now,

$\frac{{\lambda }_p}{{\lambda }_{\text{photon }}}=\sqrt{\frac{h}{2\lambda m_P{c}^2}}=\frac{1}{c}\sqrt{\frac{E}{2m_P}}$

Wavelengths associated with an electron ( \(\lambda_e\) ), proton ( \(\lambda_p\) ), and deuteron ( \(\lambda_d\) ), all moving with the same speed.

Using de Broglie's equation: \(\boldsymbol{\lambda}=\frac{\boldsymbol{h}}{\boldsymbol{m} v}\)
- Since all particles have the same speed (v), their wavelengths are inversely proportional to their masses.
- Electron has the smallest mass, so it has the longest wavelength.
- Deuteron has the highest mass, so it has the shortest wavelength.

Mass relation: \(m_e<m_p<m_d\)
thus,
wavelength relation: \(\lambda_e>\lambda_p>\lambda_d\)

• A proton (mass $m_p$​) and a photon have the same energy $E$.
• The proton moves at non-relativistic speeds.
• We need to find the ratio of the de Broglie wavelength of the proton to the wavelength of the photon.

For a photon, the energy is given by:

$$E = \frac{hc}{\lambda}$$​

Solving for $\lambda$:

$$\lambda = \frac{hc}{E}$$​

The de Broglie wavelength is given by:

$$\lambda_p = \frac{h}{p}$$​

For a non-relativistic proton, its kinetic energy is:

$$E = \frac{1}{2} m_p v^2$$

Solving for momentum $p$p:

$$p = m_p v = \sqrt{2 m_p E}$$​

Thus, the de Broglie wavelength:

$$\lambda_p = \frac{h}{\sqrt{2 m_p E}}$$​

The required ratio is:

$$\frac{\lambda_p}{\lambda} = \frac{\frac{h}{\sqrt{2 m_p E}}}{\frac{hc}{E}}$$$$= \frac{h}{\sqrt{2 m_p E}} \times \frac{E}{hc}$$=​ $$= \frac{1}{c} \times \frac{E}{\sqrt{2 m_p E}}$$​ $$= \frac{1}{c} \times \frac{\sqrt{E}}{\sqrt{2 m_p}}$$​​ $$= \frac{1}{c} \sqrt{\frac{E}{2 m_p}}$$

$\frac{1}{c}\sqrt{\frac{E}{2m_p}}$

$\frac{1}{2}{m}_p{v}^2=\frac{hc}{\lambda }=E$

$$\begin{gathered} {\lambda }_p=\frac{h}{\sqrt{2m_P\left(KE\right)}} \\ =\frac{h}{\sqrt{2m_P\left(\frac{hc}{\lambda }\right)}}=\sqrt{\frac{h\lambda }{2m_{P}c}} \end{gathered}$$

Now,

$\frac{{\lambda }_p}{{\lambda }_{\text{photon }}}=\sqrt{\frac{h}{2\lambda m_P{c}^2}}=\frac{1}{c}\sqrt{\frac{E}{2m_P}}$

$$\begin{gathered} E=\frac{hc}{\lambda } \\ E=\frac{\left(6.626\times {10}^{-34}\right)\left(3.0\times {10}^8\right)}{550\times {10}^{-9}} \\ E=\frac{1.9878\times {10}^{-25}}{550\times {10}^{-9}} \\ E=3.61\times {10}^{-19}J=2.26eV \end{gathered}$$

Cesium (Cs): Work function = 1.9 eV
Since 2.26 eV (incident photon energy) > 1.9 eV , photoelectric emission is possible.

Lithium (Li): Work function = 2.5 eV
Since 2.26 eV (incident photon energy) < 2.5 eV, photoelectric emission is not possible.

4

Energy of incoming photon

$=hv=W+K_{\max }$

The stopping potential $V_s$ is related to the maximum kinetic energy by:

$e{V}_s=K_{\max }$

Correct answer is (a).

$$\begin{gathered} \mathrm{Energy} \mathrm{associated} \mathrm{to} \mathrm{\lambda } = 550 \mathrm{nm} \mathrm{is} \\ \mathrm{E}\hspace{0.17em}= \frac{\mathrm{h} \mathrm{c}}{\mathrm{\lambda }} \\ \mathrm{E} = \frac{6.63\times {10}^{-34}\times 3\times {10}^8}{550\times {10}^{-9}}=3.61\times {10}^{-19} \mathrm{J} \\ \mathrm{E}\hspace{0.17em}= \frac{3.61\times {10}^{-19} }{1.6\times {10}^{-19} }=2.25 \mathrm{eV} \end{gathered}$$

This energy is more than the work function of cesium and less than the work fucntion of lithium. So photoelectric emission is possible for cesium only.

Wavelengths associated with an electron ( \(\lambda_e\) ), proton ( \(\lambda_p\) ), and deuteron ( \(\lambda_d\) ), all moving with the same speed.

Using de Broglie's equation: \(\boldsymbol{\lambda}=\frac{\boldsymbol{h}}{\boldsymbol{m} v}\)

- Since all particles have the same speed (v), their wavelengths are inversely proportional to their masses.

- Electron has the smallest mass, so it has the longest wavelength.

- Deuteron has the highest mass, so it has the shortest wavelength.

Mass relation: \(m_e<m_p<m_d\)

thus,

wavelength relation: \(\lambda_e>\lambda_p>\lambda_d\)

From the photoelectric equation:

$K_{\max }=hv-\varphi$

Since frequency and wavelength are related as:

$v=\frac{c}{\lambda }\Rightarrow E=\frac{hc}{\lambda }$

For the initial condition (wavelength $\lambda$ ):

$\frac{hc}{\lambda }=\varphi +eV\Rightarrow \frac{hc}{\lambda }=1+2=3eV$

For the new condition (wavelength $\frac{\lambda }{2}$ ):

$\frac{hc}{\lambda /2}=6=1+k_{\max } \therefore k_{\max }=5eV$

$$\begin{gathered} \left[E\right]=M^1{L}^2{T}^{-2} \\ \left[Pc\right]=M^1{L}^1{T}^{-1}·L^1 \\ \left[m{c}^2\right]=M^1{L}^2{T}^{-2} \\ \Rightarrow \left[E\right]=\left[pc\right]=\left[m{c}^2\right] \end{gathered}$$

From principle of homogenity $E^2=p^2{c}^2+m^2{c}^4$

Momentum is given by

$$\begin{gathered} \mathrm{p}= \frac{\mathrm{E}}{\mathrm{c}} \\ \mathrm{So}, \\ \mathrm{E} = \mathrm{p} \mathrm{c} \end{gathered}$$

$$\begin{gathered} hv=\varphi +e{V}_o \\ \Rightarrow V_0=\frac{h}{e}v-\frac{{\varphi }_0}{e} \end{gathered}$$

Graph $V_0$ vs v is linear
A is correct slope is $\frac{h}{e}$
B is wrong
& C is correct
& D is wrong
& E is correct.