NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142094
For photoelectric emission from certain metal the cut-off frequency is $v$. If radiation of frequency $2 v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be ( $\mathrm{m}$ is the electron mass)
C Given that, $v_{\mathrm{o}}=v$ From Einstein photoelectric equation, $(\mathrm{KE})_{\max }=\mathrm{h} v-\phi_{\mathrm{o}}=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}$ According to given condition, $(\mathrm{KE})_{\max } =2 \mathrm{~h} v-\mathrm{h} v_{\mathrm{o}}$ $=2 \mathrm{~h} v-\mathrm{h} v$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{h} v$ $\mathrm{v} =\sqrt{\frac{2 \mathrm{~h} v}{\mathrm{~m}}}$
MHT-CET 2013
Dual nature of radiation and Matter
142095
The ratio of moment of an electron and an $\alpha$ particle which are accelerated from rest by a potential difference of $100 \mathrm{~V}$ is
D We know that, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $q V=\frac{p^{2}}{2 m}$ $p=\sqrt{2 q m V}$ Momentum of electron $\left(\mathrm{p}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{em}_{\mathrm{e}} \mathrm{V}}$ Momentum of $\alpha$ particle $\left(\mathrm{p}_{\alpha}\right)=\sqrt{2(2 \mathrm{e}) \mathrm{m}_{\alpha} \mathrm{V}}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\alpha}}=\sqrt{\frac{2 \mathrm{em}_{\mathrm{e}} \mathrm{V}}{2(2 \mathrm{e}) \mathrm{m}_{\alpha} \mathrm{V}}}$ $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\mathrm{e}}}{2 \mathrm{~m}_{\alpha}}}$
MHT-CET 2011
Dual nature of radiation and Matter
142097
An electron is accelerated from rest to potential $V$. The final velocity of electron is
1 $\sqrt{\frac{\mathrm{eV}}{2 \mathrm{M}}}$
2 $\sqrt{\frac{4 \mathrm{eV}}{\mathrm{m}}}$
3 $\sqrt{\frac{\mathrm{eV}}{\mathrm{m}}}$
4 $\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
Explanation:
D Given that, $\mathrm{u}=0$ We know that, Change in (K.E) = work done by charge particle $\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}=\mathrm{qV} \quad(\because \mathrm{w}=\mathrm{qv})$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)=\mathrm{eV}$ $\because$ Electron is initially at rest, $\mathrm{u}=0$ $\therefore \quad \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
MHT-CET 2005
Dual nature of radiation and Matter
142098
Let the energy of an emitted photoelectron be $E$ and the wave- length of incident light be $\lambda$. What will be the change in $E$ if $\lambda$ is doubled?
1 $\mathrm{E}$
2 $\mathrm{E} / 2$
3 $2 \mathrm{E}$
4 $\mathrm{E} / 4$
Explanation:
B We know that, Energy, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ As, $\lambda^{\prime}=2 \lambda$, and $E^{\prime}=\frac{h c}{\lambda^{\prime}}$ $\mathrm{E}^{\prime}=\frac{\mathrm{hc}}{2 \lambda}$ From equation (i), putting these value, we get - $E^{\prime}=\frac{E}{2}$
142094
For photoelectric emission from certain metal the cut-off frequency is $v$. If radiation of frequency $2 v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be ( $\mathrm{m}$ is the electron mass)
C Given that, $v_{\mathrm{o}}=v$ From Einstein photoelectric equation, $(\mathrm{KE})_{\max }=\mathrm{h} v-\phi_{\mathrm{o}}=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}$ According to given condition, $(\mathrm{KE})_{\max } =2 \mathrm{~h} v-\mathrm{h} v_{\mathrm{o}}$ $=2 \mathrm{~h} v-\mathrm{h} v$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{h} v$ $\mathrm{v} =\sqrt{\frac{2 \mathrm{~h} v}{\mathrm{~m}}}$
MHT-CET 2013
Dual nature of radiation and Matter
142095
The ratio of moment of an electron and an $\alpha$ particle which are accelerated from rest by a potential difference of $100 \mathrm{~V}$ is
D We know that, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $q V=\frac{p^{2}}{2 m}$ $p=\sqrt{2 q m V}$ Momentum of electron $\left(\mathrm{p}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{em}_{\mathrm{e}} \mathrm{V}}$ Momentum of $\alpha$ particle $\left(\mathrm{p}_{\alpha}\right)=\sqrt{2(2 \mathrm{e}) \mathrm{m}_{\alpha} \mathrm{V}}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\alpha}}=\sqrt{\frac{2 \mathrm{em}_{\mathrm{e}} \mathrm{V}}{2(2 \mathrm{e}) \mathrm{m}_{\alpha} \mathrm{V}}}$ $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\mathrm{e}}}{2 \mathrm{~m}_{\alpha}}}$
MHT-CET 2011
Dual nature of radiation and Matter
142097
An electron is accelerated from rest to potential $V$. The final velocity of electron is
1 $\sqrt{\frac{\mathrm{eV}}{2 \mathrm{M}}}$
2 $\sqrt{\frac{4 \mathrm{eV}}{\mathrm{m}}}$
3 $\sqrt{\frac{\mathrm{eV}}{\mathrm{m}}}$
4 $\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
Explanation:
D Given that, $\mathrm{u}=0$ We know that, Change in (K.E) = work done by charge particle $\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}=\mathrm{qV} \quad(\because \mathrm{w}=\mathrm{qv})$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)=\mathrm{eV}$ $\because$ Electron is initially at rest, $\mathrm{u}=0$ $\therefore \quad \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
MHT-CET 2005
Dual nature of radiation and Matter
142098
Let the energy of an emitted photoelectron be $E$ and the wave- length of incident light be $\lambda$. What will be the change in $E$ if $\lambda$ is doubled?
1 $\mathrm{E}$
2 $\mathrm{E} / 2$
3 $2 \mathrm{E}$
4 $\mathrm{E} / 4$
Explanation:
B We know that, Energy, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ As, $\lambda^{\prime}=2 \lambda$, and $E^{\prime}=\frac{h c}{\lambda^{\prime}}$ $\mathrm{E}^{\prime}=\frac{\mathrm{hc}}{2 \lambda}$ From equation (i), putting these value, we get - $E^{\prime}=\frac{E}{2}$
142094
For photoelectric emission from certain metal the cut-off frequency is $v$. If radiation of frequency $2 v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be ( $\mathrm{m}$ is the electron mass)
C Given that, $v_{\mathrm{o}}=v$ From Einstein photoelectric equation, $(\mathrm{KE})_{\max }=\mathrm{h} v-\phi_{\mathrm{o}}=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}$ According to given condition, $(\mathrm{KE})_{\max } =2 \mathrm{~h} v-\mathrm{h} v_{\mathrm{o}}$ $=2 \mathrm{~h} v-\mathrm{h} v$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{h} v$ $\mathrm{v} =\sqrt{\frac{2 \mathrm{~h} v}{\mathrm{~m}}}$
MHT-CET 2013
Dual nature of radiation and Matter
142095
The ratio of moment of an electron and an $\alpha$ particle which are accelerated from rest by a potential difference of $100 \mathrm{~V}$ is
D We know that, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $q V=\frac{p^{2}}{2 m}$ $p=\sqrt{2 q m V}$ Momentum of electron $\left(\mathrm{p}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{em}_{\mathrm{e}} \mathrm{V}}$ Momentum of $\alpha$ particle $\left(\mathrm{p}_{\alpha}\right)=\sqrt{2(2 \mathrm{e}) \mathrm{m}_{\alpha} \mathrm{V}}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\alpha}}=\sqrt{\frac{2 \mathrm{em}_{\mathrm{e}} \mathrm{V}}{2(2 \mathrm{e}) \mathrm{m}_{\alpha} \mathrm{V}}}$ $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\mathrm{e}}}{2 \mathrm{~m}_{\alpha}}}$
MHT-CET 2011
Dual nature of radiation and Matter
142097
An electron is accelerated from rest to potential $V$. The final velocity of electron is
1 $\sqrt{\frac{\mathrm{eV}}{2 \mathrm{M}}}$
2 $\sqrt{\frac{4 \mathrm{eV}}{\mathrm{m}}}$
3 $\sqrt{\frac{\mathrm{eV}}{\mathrm{m}}}$
4 $\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
Explanation:
D Given that, $\mathrm{u}=0$ We know that, Change in (K.E) = work done by charge particle $\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}=\mathrm{qV} \quad(\because \mathrm{w}=\mathrm{qv})$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)=\mathrm{eV}$ $\because$ Electron is initially at rest, $\mathrm{u}=0$ $\therefore \quad \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
MHT-CET 2005
Dual nature of radiation and Matter
142098
Let the energy of an emitted photoelectron be $E$ and the wave- length of incident light be $\lambda$. What will be the change in $E$ if $\lambda$ is doubled?
1 $\mathrm{E}$
2 $\mathrm{E} / 2$
3 $2 \mathrm{E}$
4 $\mathrm{E} / 4$
Explanation:
B We know that, Energy, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ As, $\lambda^{\prime}=2 \lambda$, and $E^{\prime}=\frac{h c}{\lambda^{\prime}}$ $\mathrm{E}^{\prime}=\frac{\mathrm{hc}}{2 \lambda}$ From equation (i), putting these value, we get - $E^{\prime}=\frac{E}{2}$
142094
For photoelectric emission from certain metal the cut-off frequency is $v$. If radiation of frequency $2 v$ impinges on the metal plate, the maximum possible velocity of the emitted electron will be ( $\mathrm{m}$ is the electron mass)
C Given that, $v_{\mathrm{o}}=v$ From Einstein photoelectric equation, $(\mathrm{KE})_{\max }=\mathrm{h} v-\phi_{\mathrm{o}}=\mathrm{h} v-\mathrm{h} v_{\mathrm{o}}$ According to given condition, $(\mathrm{KE})_{\max } =2 \mathrm{~h} v-\mathrm{h} v_{\mathrm{o}}$ $=2 \mathrm{~h} v-\mathrm{h} v$ $\frac{1}{2} \mathrm{mv}^{2}=\mathrm{h} v$ $\mathrm{v} =\sqrt{\frac{2 \mathrm{~h} v}{\mathrm{~m}}}$
MHT-CET 2013
Dual nature of radiation and Matter
142095
The ratio of moment of an electron and an $\alpha$ particle which are accelerated from rest by a potential difference of $100 \mathrm{~V}$ is
D We know that, Kinetic energy $(\mathrm{KE})=\frac{1}{2} \mathrm{mv}^{2}$ $q V=\frac{p^{2}}{2 m}$ $p=\sqrt{2 q m V}$ Momentum of electron $\left(\mathrm{p}_{\mathrm{e}}\right)=\sqrt{2 \mathrm{em}_{\mathrm{e}} \mathrm{V}}$ Momentum of $\alpha$ particle $\left(\mathrm{p}_{\alpha}\right)=\sqrt{2(2 \mathrm{e}) \mathrm{m}_{\alpha} \mathrm{V}}$ On dividing equation (i) by (ii), we get - $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\alpha}}=\sqrt{\frac{2 \mathrm{em}_{\mathrm{e}} \mathrm{V}}{2(2 \mathrm{e}) \mathrm{m}_{\alpha} \mathrm{V}}}$ $\frac{\mathrm{p}_{\mathrm{e}}}{\mathrm{p}_{\alpha}}=\sqrt{\frac{\mathrm{m}_{\mathrm{e}}}{2 \mathrm{~m}_{\alpha}}}$
MHT-CET 2011
Dual nature of radiation and Matter
142097
An electron is accelerated from rest to potential $V$. The final velocity of electron is
1 $\sqrt{\frac{\mathrm{eV}}{2 \mathrm{M}}}$
2 $\sqrt{\frac{4 \mathrm{eV}}{\mathrm{m}}}$
3 $\sqrt{\frac{\mathrm{eV}}{\mathrm{m}}}$
4 $\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
Explanation:
D Given that, $\mathrm{u}=0$ We know that, Change in (K.E) = work done by charge particle $\frac{1}{2} \mathrm{mv}^{2}-\frac{1}{2} \mathrm{mu}^{2}=\mathrm{qV} \quad(\because \mathrm{w}=\mathrm{qv})$ $\frac{1}{2} \mathrm{~m}\left(\mathrm{v}^{2}-\mathrm{u}^{2}\right)=\mathrm{eV}$ $\because$ Electron is initially at rest, $\mathrm{u}=0$ $\therefore \quad \mathrm{v}=\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}$
MHT-CET 2005
Dual nature of radiation and Matter
142098
Let the energy of an emitted photoelectron be $E$ and the wave- length of incident light be $\lambda$. What will be the change in $E$ if $\lambda$ is doubled?
1 $\mathrm{E}$
2 $\mathrm{E} / 2$
3 $2 \mathrm{E}$
4 $\mathrm{E} / 4$
Explanation:
B We know that, Energy, $\mathrm{E}=\mathrm{h} v=\frac{\mathrm{hc}}{\lambda}$ As, $\lambda^{\prime}=2 \lambda$, and $E^{\prime}=\frac{h c}{\lambda^{\prime}}$ $\mathrm{E}^{\prime}=\frac{\mathrm{hc}}{2 \lambda}$ From equation (i), putting these value, we get - $E^{\prime}=\frac{E}{2}$
