142136
When two monochromatic lights of frequency, $v$ and $\frac{v}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is
1 $\frac{2}{3} v$
2 $\frac{3}{2} v$
3 $2 v$
4 $3 v$
Explanation:
B We know that, $\mathrm{E}=\phi_{\mathrm{o}}+(\mathrm{K} . \mathrm{E})_{\max }$ $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} v_{\mathrm{o}}$ Case-I $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} \frac{\mathrm{V}_{\mathrm{s}}}{2}$ Case-II $\mathrm{h} \frac{v}{2}=\mathrm{h} v_{\mathrm{o}}+\mathrm{eV} \mathrm{s}$ Multiply equation (i) by 2 and then subtract it from (ii) $2 \mathrm{~h} v-\frac{\mathrm{h} v}{2}=2 \mathrm{~h} v_{\mathrm{o}}-\mathrm{h} v_{\mathrm{o}}+2 \times \frac{\mathrm{eV}}{2}-\mathrm{eV}_{\mathrm{s}}$ $\frac{3 \mathrm{~h} v}{2}=\mathrm{h} v_{\mathrm{o}}+0$ $v_{\mathrm{o}}=\frac{3}{2} v$
Dual nature of radiation and Matter
142139
Work function of a metal is $5.2 \times 10^{-18}$, then its threshold wavelength will be
1 $736.7 \AA$
2 $760.7 \AA$
3 $301 \AA$
4 $380.4 \AA$
Explanation:
D Given that, $\phi_{\mathrm{o}}=5.2 \times 10^{-18}$ We know that, $\lambda_{\mathrm{o}}=\frac{\mathrm{hc}}{\phi_{\mathrm{o}}}$ $\lambda_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{5.2 \times 10^{-18}}$ $\lambda_{\mathrm{o}}=3.807 \times 10^{-8} \mathrm{~m}=380.7 \AA$
JIPMER-2014
Dual nature of radiation and Matter
142140
Relation between wavelength of photon and electron of same energy is
A We know that, Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ Mass of photon $=0$ Then, for the same energy the momentum of electron is more than that of photon- $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ We know that, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}} \text { and } \lambda_{\mathrm{ph}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{ph}}}$ When, $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ Then, $\quad \lambda_{\mathrm{ph}}>\lambda_{\mathrm{e}}$
JIPMER-2014
Dual nature of radiation and Matter
142141
The work function for metals $A, B$ and $C$ are respectively $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5 \mathrm{eV}$. According to Einstein's equation the metals which will emit photo, electrons for a radiation of wavelength $4100 \AA$ is / are
1 none
2 A only
3 A and B only
4 All the three metals
Explanation:
C Given that, $\lambda=4100 \AA=4100 \times 10^{-10} \mathrm{~m}$ $\phi_{\mathrm{A}}=1.92 \mathrm{eV}, \phi_{\mathrm{B}}=2.0 \mathrm{eV}, \phi_{\mathrm{C}}=5 \mathrm{eV}$ Work function for wavelength of $4100 \AA$ - $\phi =\frac{\mathrm{hc}}{\lambda}$ $\phi =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4100 \times 10^{-10}}$ $\phi =4.84 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{4.84 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =3.02 \mathrm{eV}$ Since, $\phi>\phi_{\mathrm{A}}$ and $\phi>\phi_{\mathrm{B}}$ hence only A and B will emit photoelectrons.
JIPMER-2013
Dual nature of radiation and Matter
142145
In an experimental observation of the photoelectric effect, the stopping potential was plotted against the incident light frequency as shown in the figure below: If the work function of the metal is given by $\phi_{0}$, the angle $\alpha$ is given by
4 $\alpha=\tan ^{-1}\left(\frac{\mathrm{e}}{\phi_{\mathrm{o}}}\right)$ (Here, $\mathrm{h}$ and e refer to Planck's constant and charge of electron respectively.)
Explanation:
A We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) v-\frac{\phi_{\mathrm{o}}}{\mathrm{e}}$ Equation of straight line - $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ $\therefore$ Slope $(\mathrm{m})=\frac{\mathrm{h}}{\mathrm{e}}$ $\tan \alpha=\frac{\mathrm{h}}{\mathrm{e}}$ $\alpha=\tan ^{-1}\left(\frac{\mathrm{h}}{\mathrm{e}}\right)$
142136
When two monochromatic lights of frequency, $v$ and $\frac{v}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is
1 $\frac{2}{3} v$
2 $\frac{3}{2} v$
3 $2 v$
4 $3 v$
Explanation:
B We know that, $\mathrm{E}=\phi_{\mathrm{o}}+(\mathrm{K} . \mathrm{E})_{\max }$ $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} v_{\mathrm{o}}$ Case-I $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} \frac{\mathrm{V}_{\mathrm{s}}}{2}$ Case-II $\mathrm{h} \frac{v}{2}=\mathrm{h} v_{\mathrm{o}}+\mathrm{eV} \mathrm{s}$ Multiply equation (i) by 2 and then subtract it from (ii) $2 \mathrm{~h} v-\frac{\mathrm{h} v}{2}=2 \mathrm{~h} v_{\mathrm{o}}-\mathrm{h} v_{\mathrm{o}}+2 \times \frac{\mathrm{eV}}{2}-\mathrm{eV}_{\mathrm{s}}$ $\frac{3 \mathrm{~h} v}{2}=\mathrm{h} v_{\mathrm{o}}+0$ $v_{\mathrm{o}}=\frac{3}{2} v$
Dual nature of radiation and Matter
142139
Work function of a metal is $5.2 \times 10^{-18}$, then its threshold wavelength will be
1 $736.7 \AA$
2 $760.7 \AA$
3 $301 \AA$
4 $380.4 \AA$
Explanation:
D Given that, $\phi_{\mathrm{o}}=5.2 \times 10^{-18}$ We know that, $\lambda_{\mathrm{o}}=\frac{\mathrm{hc}}{\phi_{\mathrm{o}}}$ $\lambda_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{5.2 \times 10^{-18}}$ $\lambda_{\mathrm{o}}=3.807 \times 10^{-8} \mathrm{~m}=380.7 \AA$
JIPMER-2014
Dual nature of radiation and Matter
142140
Relation between wavelength of photon and electron of same energy is
A We know that, Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ Mass of photon $=0$ Then, for the same energy the momentum of electron is more than that of photon- $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ We know that, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}} \text { and } \lambda_{\mathrm{ph}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{ph}}}$ When, $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ Then, $\quad \lambda_{\mathrm{ph}}>\lambda_{\mathrm{e}}$
JIPMER-2014
Dual nature of radiation and Matter
142141
The work function for metals $A, B$ and $C$ are respectively $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5 \mathrm{eV}$. According to Einstein's equation the metals which will emit photo, electrons for a radiation of wavelength $4100 \AA$ is / are
1 none
2 A only
3 A and B only
4 All the three metals
Explanation:
C Given that, $\lambda=4100 \AA=4100 \times 10^{-10} \mathrm{~m}$ $\phi_{\mathrm{A}}=1.92 \mathrm{eV}, \phi_{\mathrm{B}}=2.0 \mathrm{eV}, \phi_{\mathrm{C}}=5 \mathrm{eV}$ Work function for wavelength of $4100 \AA$ - $\phi =\frac{\mathrm{hc}}{\lambda}$ $\phi =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4100 \times 10^{-10}}$ $\phi =4.84 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{4.84 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =3.02 \mathrm{eV}$ Since, $\phi>\phi_{\mathrm{A}}$ and $\phi>\phi_{\mathrm{B}}$ hence only A and B will emit photoelectrons.
JIPMER-2013
Dual nature of radiation and Matter
142145
In an experimental observation of the photoelectric effect, the stopping potential was plotted against the incident light frequency as shown in the figure below: If the work function of the metal is given by $\phi_{0}$, the angle $\alpha$ is given by
4 $\alpha=\tan ^{-1}\left(\frac{\mathrm{e}}{\phi_{\mathrm{o}}}\right)$ (Here, $\mathrm{h}$ and e refer to Planck's constant and charge of electron respectively.)
Explanation:
A We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) v-\frac{\phi_{\mathrm{o}}}{\mathrm{e}}$ Equation of straight line - $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ $\therefore$ Slope $(\mathrm{m})=\frac{\mathrm{h}}{\mathrm{e}}$ $\tan \alpha=\frac{\mathrm{h}}{\mathrm{e}}$ $\alpha=\tan ^{-1}\left(\frac{\mathrm{h}}{\mathrm{e}}\right)$
142136
When two monochromatic lights of frequency, $v$ and $\frac{v}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is
1 $\frac{2}{3} v$
2 $\frac{3}{2} v$
3 $2 v$
4 $3 v$
Explanation:
B We know that, $\mathrm{E}=\phi_{\mathrm{o}}+(\mathrm{K} . \mathrm{E})_{\max }$ $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} v_{\mathrm{o}}$ Case-I $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} \frac{\mathrm{V}_{\mathrm{s}}}{2}$ Case-II $\mathrm{h} \frac{v}{2}=\mathrm{h} v_{\mathrm{o}}+\mathrm{eV} \mathrm{s}$ Multiply equation (i) by 2 and then subtract it from (ii) $2 \mathrm{~h} v-\frac{\mathrm{h} v}{2}=2 \mathrm{~h} v_{\mathrm{o}}-\mathrm{h} v_{\mathrm{o}}+2 \times \frac{\mathrm{eV}}{2}-\mathrm{eV}_{\mathrm{s}}$ $\frac{3 \mathrm{~h} v}{2}=\mathrm{h} v_{\mathrm{o}}+0$ $v_{\mathrm{o}}=\frac{3}{2} v$
Dual nature of radiation and Matter
142139
Work function of a metal is $5.2 \times 10^{-18}$, then its threshold wavelength will be
1 $736.7 \AA$
2 $760.7 \AA$
3 $301 \AA$
4 $380.4 \AA$
Explanation:
D Given that, $\phi_{\mathrm{o}}=5.2 \times 10^{-18}$ We know that, $\lambda_{\mathrm{o}}=\frac{\mathrm{hc}}{\phi_{\mathrm{o}}}$ $\lambda_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{5.2 \times 10^{-18}}$ $\lambda_{\mathrm{o}}=3.807 \times 10^{-8} \mathrm{~m}=380.7 \AA$
JIPMER-2014
Dual nature of radiation and Matter
142140
Relation between wavelength of photon and electron of same energy is
A We know that, Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ Mass of photon $=0$ Then, for the same energy the momentum of electron is more than that of photon- $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ We know that, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}} \text { and } \lambda_{\mathrm{ph}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{ph}}}$ When, $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ Then, $\quad \lambda_{\mathrm{ph}}>\lambda_{\mathrm{e}}$
JIPMER-2014
Dual nature of radiation and Matter
142141
The work function for metals $A, B$ and $C$ are respectively $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5 \mathrm{eV}$. According to Einstein's equation the metals which will emit photo, electrons for a radiation of wavelength $4100 \AA$ is / are
1 none
2 A only
3 A and B only
4 All the three metals
Explanation:
C Given that, $\lambda=4100 \AA=4100 \times 10^{-10} \mathrm{~m}$ $\phi_{\mathrm{A}}=1.92 \mathrm{eV}, \phi_{\mathrm{B}}=2.0 \mathrm{eV}, \phi_{\mathrm{C}}=5 \mathrm{eV}$ Work function for wavelength of $4100 \AA$ - $\phi =\frac{\mathrm{hc}}{\lambda}$ $\phi =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4100 \times 10^{-10}}$ $\phi =4.84 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{4.84 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =3.02 \mathrm{eV}$ Since, $\phi>\phi_{\mathrm{A}}$ and $\phi>\phi_{\mathrm{B}}$ hence only A and B will emit photoelectrons.
JIPMER-2013
Dual nature of radiation and Matter
142145
In an experimental observation of the photoelectric effect, the stopping potential was plotted against the incident light frequency as shown in the figure below: If the work function of the metal is given by $\phi_{0}$, the angle $\alpha$ is given by
4 $\alpha=\tan ^{-1}\left(\frac{\mathrm{e}}{\phi_{\mathrm{o}}}\right)$ (Here, $\mathrm{h}$ and e refer to Planck's constant and charge of electron respectively.)
Explanation:
A We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) v-\frac{\phi_{\mathrm{o}}}{\mathrm{e}}$ Equation of straight line - $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ $\therefore$ Slope $(\mathrm{m})=\frac{\mathrm{h}}{\mathrm{e}}$ $\tan \alpha=\frac{\mathrm{h}}{\mathrm{e}}$ $\alpha=\tan ^{-1}\left(\frac{\mathrm{h}}{\mathrm{e}}\right)$
142136
When two monochromatic lights of frequency, $v$ and $\frac{v}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is
1 $\frac{2}{3} v$
2 $\frac{3}{2} v$
3 $2 v$
4 $3 v$
Explanation:
B We know that, $\mathrm{E}=\phi_{\mathrm{o}}+(\mathrm{K} . \mathrm{E})_{\max }$ $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} v_{\mathrm{o}}$ Case-I $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} \frac{\mathrm{V}_{\mathrm{s}}}{2}$ Case-II $\mathrm{h} \frac{v}{2}=\mathrm{h} v_{\mathrm{o}}+\mathrm{eV} \mathrm{s}$ Multiply equation (i) by 2 and then subtract it from (ii) $2 \mathrm{~h} v-\frac{\mathrm{h} v}{2}=2 \mathrm{~h} v_{\mathrm{o}}-\mathrm{h} v_{\mathrm{o}}+2 \times \frac{\mathrm{eV}}{2}-\mathrm{eV}_{\mathrm{s}}$ $\frac{3 \mathrm{~h} v}{2}=\mathrm{h} v_{\mathrm{o}}+0$ $v_{\mathrm{o}}=\frac{3}{2} v$
Dual nature of radiation and Matter
142139
Work function of a metal is $5.2 \times 10^{-18}$, then its threshold wavelength will be
1 $736.7 \AA$
2 $760.7 \AA$
3 $301 \AA$
4 $380.4 \AA$
Explanation:
D Given that, $\phi_{\mathrm{o}}=5.2 \times 10^{-18}$ We know that, $\lambda_{\mathrm{o}}=\frac{\mathrm{hc}}{\phi_{\mathrm{o}}}$ $\lambda_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{5.2 \times 10^{-18}}$ $\lambda_{\mathrm{o}}=3.807 \times 10^{-8} \mathrm{~m}=380.7 \AA$
JIPMER-2014
Dual nature of radiation and Matter
142140
Relation between wavelength of photon and electron of same energy is
A We know that, Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ Mass of photon $=0$ Then, for the same energy the momentum of electron is more than that of photon- $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ We know that, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}} \text { and } \lambda_{\mathrm{ph}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{ph}}}$ When, $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ Then, $\quad \lambda_{\mathrm{ph}}>\lambda_{\mathrm{e}}$
JIPMER-2014
Dual nature of radiation and Matter
142141
The work function for metals $A, B$ and $C$ are respectively $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5 \mathrm{eV}$. According to Einstein's equation the metals which will emit photo, electrons for a radiation of wavelength $4100 \AA$ is / are
1 none
2 A only
3 A and B only
4 All the three metals
Explanation:
C Given that, $\lambda=4100 \AA=4100 \times 10^{-10} \mathrm{~m}$ $\phi_{\mathrm{A}}=1.92 \mathrm{eV}, \phi_{\mathrm{B}}=2.0 \mathrm{eV}, \phi_{\mathrm{C}}=5 \mathrm{eV}$ Work function for wavelength of $4100 \AA$ - $\phi =\frac{\mathrm{hc}}{\lambda}$ $\phi =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4100 \times 10^{-10}}$ $\phi =4.84 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{4.84 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =3.02 \mathrm{eV}$ Since, $\phi>\phi_{\mathrm{A}}$ and $\phi>\phi_{\mathrm{B}}$ hence only A and B will emit photoelectrons.
JIPMER-2013
Dual nature of radiation and Matter
142145
In an experimental observation of the photoelectric effect, the stopping potential was plotted against the incident light frequency as shown in the figure below: If the work function of the metal is given by $\phi_{0}$, the angle $\alpha$ is given by
4 $\alpha=\tan ^{-1}\left(\frac{\mathrm{e}}{\phi_{\mathrm{o}}}\right)$ (Here, $\mathrm{h}$ and e refer to Planck's constant and charge of electron respectively.)
Explanation:
A We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) v-\frac{\phi_{\mathrm{o}}}{\mathrm{e}}$ Equation of straight line - $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ $\therefore$ Slope $(\mathrm{m})=\frac{\mathrm{h}}{\mathrm{e}}$ $\tan \alpha=\frac{\mathrm{h}}{\mathrm{e}}$ $\alpha=\tan ^{-1}\left(\frac{\mathrm{h}}{\mathrm{e}}\right)$
142136
When two monochromatic lights of frequency, $v$ and $\frac{v}{2}$ are incident on a photoelectric metal, their stopping potential becomes $\frac{V_{s}}{2}$ and $V_{s}$ respectively. The threshold frequency for this metal is
1 $\frac{2}{3} v$
2 $\frac{3}{2} v$
3 $2 v$
4 $3 v$
Explanation:
B We know that, $\mathrm{E}=\phi_{\mathrm{o}}+(\mathrm{K} . \mathrm{E})_{\max }$ $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} v_{\mathrm{o}}$ Case-I $\mathrm{h} v=\mathrm{h} v_{\mathrm{o}}+\mathrm{e} \frac{\mathrm{V}_{\mathrm{s}}}{2}$ Case-II $\mathrm{h} \frac{v}{2}=\mathrm{h} v_{\mathrm{o}}+\mathrm{eV} \mathrm{s}$ Multiply equation (i) by 2 and then subtract it from (ii) $2 \mathrm{~h} v-\frac{\mathrm{h} v}{2}=2 \mathrm{~h} v_{\mathrm{o}}-\mathrm{h} v_{\mathrm{o}}+2 \times \frac{\mathrm{eV}}{2}-\mathrm{eV}_{\mathrm{s}}$ $\frac{3 \mathrm{~h} v}{2}=\mathrm{h} v_{\mathrm{o}}+0$ $v_{\mathrm{o}}=\frac{3}{2} v$
Dual nature of radiation and Matter
142139
Work function of a metal is $5.2 \times 10^{-18}$, then its threshold wavelength will be
1 $736.7 \AA$
2 $760.7 \AA$
3 $301 \AA$
4 $380.4 \AA$
Explanation:
D Given that, $\phi_{\mathrm{o}}=5.2 \times 10^{-18}$ We know that, $\lambda_{\mathrm{o}}=\frac{\mathrm{hc}}{\phi_{\mathrm{o}}}$ $\lambda_{\mathrm{o}}=\frac{6.6 \times 10^{-34} \times 3 \times 10^{8}}{5.2 \times 10^{-18}}$ $\lambda_{\mathrm{o}}=3.807 \times 10^{-8} \mathrm{~m}=380.7 \AA$
JIPMER-2014
Dual nature of radiation and Matter
142140
Relation between wavelength of photon and electron of same energy is
A We know that, Mass of electron $=9.1 \times 10^{-31} \mathrm{~kg}$ Mass of photon $=0$ Then, for the same energy the momentum of electron is more than that of photon- $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ We know that, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ $\lambda_{\mathrm{e}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{e}}} \text { and } \lambda_{\mathrm{ph}}=\frac{\mathrm{h}}{\mathrm{p}_{\mathrm{ph}}}$ When, $\mathrm{p}_{\mathrm{e}}>\mathrm{p}_{\mathrm{ph}}$ Then, $\quad \lambda_{\mathrm{ph}}>\lambda_{\mathrm{e}}$
JIPMER-2014
Dual nature of radiation and Matter
142141
The work function for metals $A, B$ and $C$ are respectively $1.92 \mathrm{eV}, 2.0 \mathrm{eV}$ and $5 \mathrm{eV}$. According to Einstein's equation the metals which will emit photo, electrons for a radiation of wavelength $4100 \AA$ is / are
1 none
2 A only
3 A and B only
4 All the three metals
Explanation:
C Given that, $\lambda=4100 \AA=4100 \times 10^{-10} \mathrm{~m}$ $\phi_{\mathrm{A}}=1.92 \mathrm{eV}, \phi_{\mathrm{B}}=2.0 \mathrm{eV}, \phi_{\mathrm{C}}=5 \mathrm{eV}$ Work function for wavelength of $4100 \AA$ - $\phi =\frac{\mathrm{hc}}{\lambda}$ $\phi =\frac{6.62 \times 10^{-34} \times 3 \times 10^{8}}{4100 \times 10^{-10}}$ $\phi =4.84 \times 10^{-19} \mathrm{~J}$ $\phi =\frac{4.84 \times 10^{-19}}{1.6 \times 10^{-19}} \mathrm{eV}$ $\phi =3.02 \mathrm{eV}$ Since, $\phi>\phi_{\mathrm{A}}$ and $\phi>\phi_{\mathrm{B}}$ hence only A and B will emit photoelectrons.
JIPMER-2013
Dual nature of radiation and Matter
142145
In an experimental observation of the photoelectric effect, the stopping potential was plotted against the incident light frequency as shown in the figure below: If the work function of the metal is given by $\phi_{0}$, the angle $\alpha$ is given by
4 $\alpha=\tan ^{-1}\left(\frac{\mathrm{e}}{\phi_{\mathrm{o}}}\right)$ (Here, $\mathrm{h}$ and e refer to Planck's constant and charge of electron respectively.)
Explanation:
A We know that, $\mathrm{eV}_{\mathrm{o}}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{V}_{\mathrm{o}}=\left(\frac{\mathrm{h}}{\mathrm{e}}\right) v-\frac{\phi_{\mathrm{o}}}{\mathrm{e}}$ Equation of straight line - $\mathrm{y}=\mathrm{mx}+\mathrm{c}$ $\therefore$ Slope $(\mathrm{m})=\frac{\mathrm{h}}{\mathrm{e}}$ $\tan \alpha=\frac{\mathrm{h}}{\mathrm{e}}$ $\alpha=\tan ^{-1}\left(\frac{\mathrm{h}}{\mathrm{e}}\right)$
