142485
If the energy of the photon is increased by a factor of 4, then its momentum:
1 does not change
2 decreases by a factor of 4
3 increases by a factor of 4
4 decreases by a factor of 2
Explanation:
C Momentum of photon, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Energy of photon $(E)=\frac{h c}{\lambda}$ From equation (i) and equation (ii) $\therefore \quad \mathrm{E}=\mathrm{pc}$ So, if the energy of the photon is increased by a factor of 4 then its momentum increases by a factor of 4 .
UPSEE - 2004
Dual nature of radiation and Matter
142486
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential. Then, the ratio of their de-Broglie wavelengths is
1 1
2 $\sqrt{\frac{m_{e}}{m_{p}}}$
3 $\frac{m_{e}}{m_{p}}$
4 $\frac{m_{p}}{m_{e}}$
5 $\sqrt{\frac{m_{p}}{m_{e}}}$
Explanation:
E Given that, Mass of electron $=\mathrm{m}_{\mathrm{e}}$ Mass of proton $=\mathrm{m}_{\mathrm{p}}$ We know that, de-Broglie wavelength - $\lambda=\frac{\mathrm{h}}{\mathrm{p}} {[\because \mathrm{p}=\mathrm{mv}]}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \ldots \text { (i) }$ The kinetic energy of electron $\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}^{2}=\frac{2 \mathrm{k}}{\mathrm{m}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}$ Putting the value of $v$ from equation (ii) in equation (i) Then, $\lambda=\frac{\mathrm{h}}{\mathrm{m} \cdot \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda_{\mathrm{e}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}$ $\lambda_{\mathrm{p}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}$ Equation (iii) divided by equation (iv), we get - $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}}{\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}}$
Kerala CEE- 2014 VITEEE-2007
Dual nature of radiation and Matter
142487
The de-Broglie wavelength and kinetic energy of a particle is $2000 \AA$ and $1 \mathrm{eV}$ respectively. If its kinetic energy becomes $1 \mathrm{MeV}$, then its deBroglie wavelength is
1 $2 \AA$
2 $1 \AA$
3 $4 \AA$
4 $10 \AA$
5 $5 \AA$
Explanation:
A Given that, $\lambda_{1}=2000 \AA$, $\mathrm{K}_{1}=1 \mathrm{eV}, \mathrm{K}_{2}=$ $1 \mathrm{MeV}=1 \times 10^{6} \mathrm{eV}$ We know that, de-Broglie wavelength of a particle- Wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\because$ Mass of the particle remains constant So, $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$ $\lambda \propto \sqrt{\frac{1}{\mathrm{~K}}}$ $\frac{2000}{\lambda_{2}}=\sqrt{\frac{1 \times 10^{6} \mathrm{eV}}{1 \mathrm{eV}}}$ $\therefore \quad \lambda_{2}=\frac{2000}{10^{3}}=2 \AA$
Kerala CEE- 2013
Dual nature of radiation and Matter
142488
The work functions of two metals are $2.75 \mathrm{eV}$, and $2 \mathrm{eV}$ respectively. If these are irradiated by photons of energy $3 \mathrm{eV}$, the ratio of maximum momenta of the photoelectrons emitted respectively by them is
1 $1: 2$
2 $1: 3$
3 $1: 4$
4 $2: 1$
5 $4: 1$
Explanation:
A Given that, $\phi_{1}=2.25 \mathrm{eV}, \phi_{2}=2.25 \mathrm{eV}$ Energy of photon, $h v=3 \mathrm{eV}$ We know that, Max. momentum $(\mathrm{p})=\sqrt{2 \mathrm{~m}(\mathrm{KE})}$ Ratio of maximum momentum $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{\mathrm{h} v-\phi_{1}}{\mathrm{~h} v-\phi_{2}}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{3-2.75}{3-2}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{0.25}{1}}=\frac{0.5}{1}=\frac{5}{10}=\frac{1}{2}$ $\therefore \quad \mathrm{p}_{1} : \mathrm{p}_{2}=1: 2$
142485
If the energy of the photon is increased by a factor of 4, then its momentum:
1 does not change
2 decreases by a factor of 4
3 increases by a factor of 4
4 decreases by a factor of 2
Explanation:
C Momentum of photon, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Energy of photon $(E)=\frac{h c}{\lambda}$ From equation (i) and equation (ii) $\therefore \quad \mathrm{E}=\mathrm{pc}$ So, if the energy of the photon is increased by a factor of 4 then its momentum increases by a factor of 4 .
UPSEE - 2004
Dual nature of radiation and Matter
142486
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential. Then, the ratio of their de-Broglie wavelengths is
1 1
2 $\sqrt{\frac{m_{e}}{m_{p}}}$
3 $\frac{m_{e}}{m_{p}}$
4 $\frac{m_{p}}{m_{e}}$
5 $\sqrt{\frac{m_{p}}{m_{e}}}$
Explanation:
E Given that, Mass of electron $=\mathrm{m}_{\mathrm{e}}$ Mass of proton $=\mathrm{m}_{\mathrm{p}}$ We know that, de-Broglie wavelength - $\lambda=\frac{\mathrm{h}}{\mathrm{p}} {[\because \mathrm{p}=\mathrm{mv}]}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \ldots \text { (i) }$ The kinetic energy of electron $\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}^{2}=\frac{2 \mathrm{k}}{\mathrm{m}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}$ Putting the value of $v$ from equation (ii) in equation (i) Then, $\lambda=\frac{\mathrm{h}}{\mathrm{m} \cdot \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda_{\mathrm{e}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}$ $\lambda_{\mathrm{p}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}$ Equation (iii) divided by equation (iv), we get - $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}}{\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}}$
Kerala CEE- 2014 VITEEE-2007
Dual nature of radiation and Matter
142487
The de-Broglie wavelength and kinetic energy of a particle is $2000 \AA$ and $1 \mathrm{eV}$ respectively. If its kinetic energy becomes $1 \mathrm{MeV}$, then its deBroglie wavelength is
1 $2 \AA$
2 $1 \AA$
3 $4 \AA$
4 $10 \AA$
5 $5 \AA$
Explanation:
A Given that, $\lambda_{1}=2000 \AA$, $\mathrm{K}_{1}=1 \mathrm{eV}, \mathrm{K}_{2}=$ $1 \mathrm{MeV}=1 \times 10^{6} \mathrm{eV}$ We know that, de-Broglie wavelength of a particle- Wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\because$ Mass of the particle remains constant So, $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$ $\lambda \propto \sqrt{\frac{1}{\mathrm{~K}}}$ $\frac{2000}{\lambda_{2}}=\sqrt{\frac{1 \times 10^{6} \mathrm{eV}}{1 \mathrm{eV}}}$ $\therefore \quad \lambda_{2}=\frac{2000}{10^{3}}=2 \AA$
Kerala CEE- 2013
Dual nature of radiation and Matter
142488
The work functions of two metals are $2.75 \mathrm{eV}$, and $2 \mathrm{eV}$ respectively. If these are irradiated by photons of energy $3 \mathrm{eV}$, the ratio of maximum momenta of the photoelectrons emitted respectively by them is
1 $1: 2$
2 $1: 3$
3 $1: 4$
4 $2: 1$
5 $4: 1$
Explanation:
A Given that, $\phi_{1}=2.25 \mathrm{eV}, \phi_{2}=2.25 \mathrm{eV}$ Energy of photon, $h v=3 \mathrm{eV}$ We know that, Max. momentum $(\mathrm{p})=\sqrt{2 \mathrm{~m}(\mathrm{KE})}$ Ratio of maximum momentum $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{\mathrm{h} v-\phi_{1}}{\mathrm{~h} v-\phi_{2}}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{3-2.75}{3-2}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{0.25}{1}}=\frac{0.5}{1}=\frac{5}{10}=\frac{1}{2}$ $\therefore \quad \mathrm{p}_{1} : \mathrm{p}_{2}=1: 2$
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Dual nature of radiation and Matter
142485
If the energy of the photon is increased by a factor of 4, then its momentum:
1 does not change
2 decreases by a factor of 4
3 increases by a factor of 4
4 decreases by a factor of 2
Explanation:
C Momentum of photon, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Energy of photon $(E)=\frac{h c}{\lambda}$ From equation (i) and equation (ii) $\therefore \quad \mathrm{E}=\mathrm{pc}$ So, if the energy of the photon is increased by a factor of 4 then its momentum increases by a factor of 4 .
UPSEE - 2004
Dual nature of radiation and Matter
142486
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential. Then, the ratio of their de-Broglie wavelengths is
1 1
2 $\sqrt{\frac{m_{e}}{m_{p}}}$
3 $\frac{m_{e}}{m_{p}}$
4 $\frac{m_{p}}{m_{e}}$
5 $\sqrt{\frac{m_{p}}{m_{e}}}$
Explanation:
E Given that, Mass of electron $=\mathrm{m}_{\mathrm{e}}$ Mass of proton $=\mathrm{m}_{\mathrm{p}}$ We know that, de-Broglie wavelength - $\lambda=\frac{\mathrm{h}}{\mathrm{p}} {[\because \mathrm{p}=\mathrm{mv}]}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \ldots \text { (i) }$ The kinetic energy of electron $\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}^{2}=\frac{2 \mathrm{k}}{\mathrm{m}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}$ Putting the value of $v$ from equation (ii) in equation (i) Then, $\lambda=\frac{\mathrm{h}}{\mathrm{m} \cdot \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda_{\mathrm{e}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}$ $\lambda_{\mathrm{p}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}$ Equation (iii) divided by equation (iv), we get - $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}}{\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}}$
Kerala CEE- 2014 VITEEE-2007
Dual nature of radiation and Matter
142487
The de-Broglie wavelength and kinetic energy of a particle is $2000 \AA$ and $1 \mathrm{eV}$ respectively. If its kinetic energy becomes $1 \mathrm{MeV}$, then its deBroglie wavelength is
1 $2 \AA$
2 $1 \AA$
3 $4 \AA$
4 $10 \AA$
5 $5 \AA$
Explanation:
A Given that, $\lambda_{1}=2000 \AA$, $\mathrm{K}_{1}=1 \mathrm{eV}, \mathrm{K}_{2}=$ $1 \mathrm{MeV}=1 \times 10^{6} \mathrm{eV}$ We know that, de-Broglie wavelength of a particle- Wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\because$ Mass of the particle remains constant So, $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$ $\lambda \propto \sqrt{\frac{1}{\mathrm{~K}}}$ $\frac{2000}{\lambda_{2}}=\sqrt{\frac{1 \times 10^{6} \mathrm{eV}}{1 \mathrm{eV}}}$ $\therefore \quad \lambda_{2}=\frac{2000}{10^{3}}=2 \AA$
Kerala CEE- 2013
Dual nature of radiation and Matter
142488
The work functions of two metals are $2.75 \mathrm{eV}$, and $2 \mathrm{eV}$ respectively. If these are irradiated by photons of energy $3 \mathrm{eV}$, the ratio of maximum momenta of the photoelectrons emitted respectively by them is
1 $1: 2$
2 $1: 3$
3 $1: 4$
4 $2: 1$
5 $4: 1$
Explanation:
A Given that, $\phi_{1}=2.25 \mathrm{eV}, \phi_{2}=2.25 \mathrm{eV}$ Energy of photon, $h v=3 \mathrm{eV}$ We know that, Max. momentum $(\mathrm{p})=\sqrt{2 \mathrm{~m}(\mathrm{KE})}$ Ratio of maximum momentum $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{\mathrm{h} v-\phi_{1}}{\mathrm{~h} v-\phi_{2}}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{3-2.75}{3-2}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{0.25}{1}}=\frac{0.5}{1}=\frac{5}{10}=\frac{1}{2}$ $\therefore \quad \mathrm{p}_{1} : \mathrm{p}_{2}=1: 2$
142485
If the energy of the photon is increased by a factor of 4, then its momentum:
1 does not change
2 decreases by a factor of 4
3 increases by a factor of 4
4 decreases by a factor of 2
Explanation:
C Momentum of photon, $\mathrm{p}=\frac{\mathrm{h}}{\lambda}$ Energy of photon $(E)=\frac{h c}{\lambda}$ From equation (i) and equation (ii) $\therefore \quad \mathrm{E}=\mathrm{pc}$ So, if the energy of the photon is increased by a factor of 4 then its momentum increases by a factor of 4 .
UPSEE - 2004
Dual nature of radiation and Matter
142486
An electron of mass $m_{e}$ and a proton of mass $m_{p}$ are accelerated through the same potential. Then, the ratio of their de-Broglie wavelengths is
1 1
2 $\sqrt{\frac{m_{e}}{m_{p}}}$
3 $\frac{m_{e}}{m_{p}}$
4 $\frac{m_{p}}{m_{e}}$
5 $\sqrt{\frac{m_{p}}{m_{e}}}$
Explanation:
E Given that, Mass of electron $=\mathrm{m}_{\mathrm{e}}$ Mass of proton $=\mathrm{m}_{\mathrm{p}}$ We know that, de-Broglie wavelength - $\lambda=\frac{\mathrm{h}}{\mathrm{p}} {[\because \mathrm{p}=\mathrm{mv}]}$ $\lambda=\frac{\mathrm{h}}{\mathrm{mv}} \ldots \text { (i) }$ The kinetic energy of electron $\mathrm{k}=\frac{1}{2} \mathrm{mv}^{2}$ $\mathrm{v}^{2}=\frac{2 \mathrm{k}}{\mathrm{m}}$ $\mathrm{v}=\sqrt{\frac{2 \mathrm{k}}{\mathrm{m}}}$ Putting the value of $v$ from equation (ii) in equation (i) Then, $\lambda=\frac{\mathrm{h}}{\mathrm{m} \cdot \sqrt{\frac{2 \mathrm{~K}}{\mathrm{~m}}}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\lambda \propto \frac{1}{\sqrt{\mathrm{m}}}$ $\lambda_{\mathrm{e}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}$ $\lambda_{\mathrm{p}}=\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}$ Equation (iii) divided by equation (iv), we get - $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\frac{\frac{1}{\sqrt{\mathrm{m}_{\mathrm{e}}}}}{\frac{1}{\sqrt{\mathrm{m}_{\mathrm{p}}}}}$ $\frac{\lambda_{\mathrm{e}}}{\lambda_{\mathrm{p}}}=\sqrt{\frac{\mathrm{m}_{\mathrm{p}}}{\mathrm{m}_{\mathrm{e}}}}$
Kerala CEE- 2014 VITEEE-2007
Dual nature of radiation and Matter
142487
The de-Broglie wavelength and kinetic energy of a particle is $2000 \AA$ and $1 \mathrm{eV}$ respectively. If its kinetic energy becomes $1 \mathrm{MeV}$, then its deBroglie wavelength is
1 $2 \AA$
2 $1 \AA$
3 $4 \AA$
4 $10 \AA$
5 $5 \AA$
Explanation:
A Given that, $\lambda_{1}=2000 \AA$, $\mathrm{K}_{1}=1 \mathrm{eV}, \mathrm{K}_{2}=$ $1 \mathrm{MeV}=1 \times 10^{6} \mathrm{eV}$ We know that, de-Broglie wavelength of a particle- Wavelength, $\lambda=\frac{\mathrm{h}}{\mathrm{p}}=\frac{\mathrm{h}}{\sqrt{2 \mathrm{mK}}}$ $\because$ Mass of the particle remains constant So, $\therefore \quad \frac{\lambda_{1}}{\lambda_{2}}=\sqrt{\frac{\mathrm{K}_{2}}{\mathrm{~K}_{1}}}$ $\lambda \propto \sqrt{\frac{1}{\mathrm{~K}}}$ $\frac{2000}{\lambda_{2}}=\sqrt{\frac{1 \times 10^{6} \mathrm{eV}}{1 \mathrm{eV}}}$ $\therefore \quad \lambda_{2}=\frac{2000}{10^{3}}=2 \AA$
Kerala CEE- 2013
Dual nature of radiation and Matter
142488
The work functions of two metals are $2.75 \mathrm{eV}$, and $2 \mathrm{eV}$ respectively. If these are irradiated by photons of energy $3 \mathrm{eV}$, the ratio of maximum momenta of the photoelectrons emitted respectively by them is
1 $1: 2$
2 $1: 3$
3 $1: 4$
4 $2: 1$
5 $4: 1$
Explanation:
A Given that, $\phi_{1}=2.25 \mathrm{eV}, \phi_{2}=2.25 \mathrm{eV}$ Energy of photon, $h v=3 \mathrm{eV}$ We know that, Max. momentum $(\mathrm{p})=\sqrt{2 \mathrm{~m}(\mathrm{KE})}$ Ratio of maximum momentum $\therefore \quad \frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{\mathrm{h} v-\phi_{1}}{\mathrm{~h} v-\phi_{2}}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{3-2.75}{3-2}}$ $\frac{\mathrm{p}_{1}}{\mathrm{p}_{2}} =\sqrt{\frac{0.25}{1}}=\frac{0.5}{1}=\frac{5}{10}=\frac{1}{2}$ $\therefore \quad \mathrm{p}_{1} : \mathrm{p}_{2}=1: 2$