141983
The kinetic energy of emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy, the incident light must have wavelength:
B From Einstein photoelectric equation $(\mathrm{K} . \mathrm{E})_{\max }=\mathrm{E}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $2 \mathrm{E}=-\phi_{\mathrm{o}}$ Subtract equation (ii) from (i), we get- $2 \mathrm{E}-\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\frac{\mathrm{hc}}{\lambda}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=\mathrm{E}+\frac{\mathrm{hc}}{\lambda}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=\frac{\mathrm{E} \lambda+\mathrm{hc}}{\lambda}$ $\lambda^{\prime}=\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$
Shift-I
Dual nature of radiation and Matter
141985
Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$, respectively are incident on a metallic surface. If $\lambda_{1}=3 \lambda_{2}$ then:
1 $K_{1}>\frac{K_{2}}{3}$
2 $\mathrm{K}_{1} \lt \frac{\mathrm{K}_{2}}{3}$
3 $\mathrm{K}_{1}=\frac{\mathrm{K}_{2}}{3}$
4 $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{3}$
Explanation:
B Given that, $\lambda_{1}=3 \lambda_{2}$ We know that $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Case I $\mathrm{K}_{1}=\frac{\mathrm{hc}}{\lambda_{1}}-\phi_{\mathrm{o}}$ Case II $\mathrm{K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Then from equation (i), we get- $\mathrm{K}_{1}=\frac{\mathrm{hc}}{3 \lambda_{2}}-\phi_{\mathrm{o}}$ $3 \mathrm{~K}_{1}=\frac{\mathrm{hc}}{\lambda_{2}}-3 \phi_{\mathrm{o}}$ From equation (ii) and (iii), we get- $3 \mathrm{~K}_{1}=\mathrm{K}_{2}-2 \phi_{\mathrm{o}}$ $3 \mathrm{~K}_{1} \lt \mathrm{K}_{2}$ $\mathrm{~K}_{1} \lt \frac{\mathrm{K}_{2}}{3}$
JEE Main-28.06.2022
Dual nature of radiation and Matter
141986
A parallel beam of light of wavelength $900 \mathrm{~nm}$ and intensity $100 \mathrm{Wm}^{-2}$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \mathrm{~cm}^{2}$ area perpendicular to the beam in one second is:
1 $3 \times 10^{16}$
2 $4.5 \times 10^{16}$
3 $4.5 \times 10^{17}$
4 $4.5 \times 10^{20}$
Explanation:
B Given that, $\lambda=900 \mathrm{~nm}=9 \times 10^{-7} \mathrm{~m}$ $\mathrm{I}=100 \mathrm{~W} / \mathrm{m}^{2}$ $\mathrm{~A}=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}$ Number of photons in $1 \mathrm{sec}$ is given by $\mathrm{n}=\frac{\mathrm{E}_{\text {net }}}{\mathrm{E}_{\text {single photon }}}=\frac{\lambda \mathrm{AI}}{\mathrm{hc}}$ $\mathrm{n}=\frac{9 \times 10^{-7} \times 10^{-4} \times 100}{6.62 \times 10^{-34} \times 3 \times 10^{8}}$ $\mathrm{n}=4.5 \times 10^{16}$
JEE Main-26.07.2022
Dual nature of radiation and Matter
141987
Photoelectric emission from a metal begins at a frequency of $6 \times 10^{14} \mathrm{~Hz}$. The emitted electrons are fully stopped by a retarding potential of 3.3 $V$. What will be the wavelength of the incident radiation?
141983
The kinetic energy of emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy, the incident light must have wavelength:
B From Einstein photoelectric equation $(\mathrm{K} . \mathrm{E})_{\max }=\mathrm{E}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $2 \mathrm{E}=-\phi_{\mathrm{o}}$ Subtract equation (ii) from (i), we get- $2 \mathrm{E}-\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\frac{\mathrm{hc}}{\lambda}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=\mathrm{E}+\frac{\mathrm{hc}}{\lambda}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=\frac{\mathrm{E} \lambda+\mathrm{hc}}{\lambda}$ $\lambda^{\prime}=\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$
Shift-I
Dual nature of radiation and Matter
141985
Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$, respectively are incident on a metallic surface. If $\lambda_{1}=3 \lambda_{2}$ then:
1 $K_{1}>\frac{K_{2}}{3}$
2 $\mathrm{K}_{1} \lt \frac{\mathrm{K}_{2}}{3}$
3 $\mathrm{K}_{1}=\frac{\mathrm{K}_{2}}{3}$
4 $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{3}$
Explanation:
B Given that, $\lambda_{1}=3 \lambda_{2}$ We know that $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Case I $\mathrm{K}_{1}=\frac{\mathrm{hc}}{\lambda_{1}}-\phi_{\mathrm{o}}$ Case II $\mathrm{K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Then from equation (i), we get- $\mathrm{K}_{1}=\frac{\mathrm{hc}}{3 \lambda_{2}}-\phi_{\mathrm{o}}$ $3 \mathrm{~K}_{1}=\frac{\mathrm{hc}}{\lambda_{2}}-3 \phi_{\mathrm{o}}$ From equation (ii) and (iii), we get- $3 \mathrm{~K}_{1}=\mathrm{K}_{2}-2 \phi_{\mathrm{o}}$ $3 \mathrm{~K}_{1} \lt \mathrm{K}_{2}$ $\mathrm{~K}_{1} \lt \frac{\mathrm{K}_{2}}{3}$
JEE Main-28.06.2022
Dual nature of radiation and Matter
141986
A parallel beam of light of wavelength $900 \mathrm{~nm}$ and intensity $100 \mathrm{Wm}^{-2}$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \mathrm{~cm}^{2}$ area perpendicular to the beam in one second is:
1 $3 \times 10^{16}$
2 $4.5 \times 10^{16}$
3 $4.5 \times 10^{17}$
4 $4.5 \times 10^{20}$
Explanation:
B Given that, $\lambda=900 \mathrm{~nm}=9 \times 10^{-7} \mathrm{~m}$ $\mathrm{I}=100 \mathrm{~W} / \mathrm{m}^{2}$ $\mathrm{~A}=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}$ Number of photons in $1 \mathrm{sec}$ is given by $\mathrm{n}=\frac{\mathrm{E}_{\text {net }}}{\mathrm{E}_{\text {single photon }}}=\frac{\lambda \mathrm{AI}}{\mathrm{hc}}$ $\mathrm{n}=\frac{9 \times 10^{-7} \times 10^{-4} \times 100}{6.62 \times 10^{-34} \times 3 \times 10^{8}}$ $\mathrm{n}=4.5 \times 10^{16}$
JEE Main-26.07.2022
Dual nature of radiation and Matter
141987
Photoelectric emission from a metal begins at a frequency of $6 \times 10^{14} \mathrm{~Hz}$. The emitted electrons are fully stopped by a retarding potential of 3.3 $V$. What will be the wavelength of the incident radiation?
141983
The kinetic energy of emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy, the incident light must have wavelength:
B From Einstein photoelectric equation $(\mathrm{K} . \mathrm{E})_{\max }=\mathrm{E}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $2 \mathrm{E}=-\phi_{\mathrm{o}}$ Subtract equation (ii) from (i), we get- $2 \mathrm{E}-\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\frac{\mathrm{hc}}{\lambda}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=\mathrm{E}+\frac{\mathrm{hc}}{\lambda}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=\frac{\mathrm{E} \lambda+\mathrm{hc}}{\lambda}$ $\lambda^{\prime}=\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$
Shift-I
Dual nature of radiation and Matter
141985
Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$, respectively are incident on a metallic surface. If $\lambda_{1}=3 \lambda_{2}$ then:
1 $K_{1}>\frac{K_{2}}{3}$
2 $\mathrm{K}_{1} \lt \frac{\mathrm{K}_{2}}{3}$
3 $\mathrm{K}_{1}=\frac{\mathrm{K}_{2}}{3}$
4 $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{3}$
Explanation:
B Given that, $\lambda_{1}=3 \lambda_{2}$ We know that $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Case I $\mathrm{K}_{1}=\frac{\mathrm{hc}}{\lambda_{1}}-\phi_{\mathrm{o}}$ Case II $\mathrm{K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Then from equation (i), we get- $\mathrm{K}_{1}=\frac{\mathrm{hc}}{3 \lambda_{2}}-\phi_{\mathrm{o}}$ $3 \mathrm{~K}_{1}=\frac{\mathrm{hc}}{\lambda_{2}}-3 \phi_{\mathrm{o}}$ From equation (ii) and (iii), we get- $3 \mathrm{~K}_{1}=\mathrm{K}_{2}-2 \phi_{\mathrm{o}}$ $3 \mathrm{~K}_{1} \lt \mathrm{K}_{2}$ $\mathrm{~K}_{1} \lt \frac{\mathrm{K}_{2}}{3}$
JEE Main-28.06.2022
Dual nature of radiation and Matter
141986
A parallel beam of light of wavelength $900 \mathrm{~nm}$ and intensity $100 \mathrm{Wm}^{-2}$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \mathrm{~cm}^{2}$ area perpendicular to the beam in one second is:
1 $3 \times 10^{16}$
2 $4.5 \times 10^{16}$
3 $4.5 \times 10^{17}$
4 $4.5 \times 10^{20}$
Explanation:
B Given that, $\lambda=900 \mathrm{~nm}=9 \times 10^{-7} \mathrm{~m}$ $\mathrm{I}=100 \mathrm{~W} / \mathrm{m}^{2}$ $\mathrm{~A}=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}$ Number of photons in $1 \mathrm{sec}$ is given by $\mathrm{n}=\frac{\mathrm{E}_{\text {net }}}{\mathrm{E}_{\text {single photon }}}=\frac{\lambda \mathrm{AI}}{\mathrm{hc}}$ $\mathrm{n}=\frac{9 \times 10^{-7} \times 10^{-4} \times 100}{6.62 \times 10^{-34} \times 3 \times 10^{8}}$ $\mathrm{n}=4.5 \times 10^{16}$
JEE Main-26.07.2022
Dual nature of radiation and Matter
141987
Photoelectric emission from a metal begins at a frequency of $6 \times 10^{14} \mathrm{~Hz}$. The emitted electrons are fully stopped by a retarding potential of 3.3 $V$. What will be the wavelength of the incident radiation?
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
141983
The kinetic energy of emitted electron is $E$ when the light incident on the metal has wavelength $\lambda$. To double the kinetic energy, the incident light must have wavelength:
B From Einstein photoelectric equation $(\mathrm{K} . \mathrm{E})_{\max }=\mathrm{E}=\mathrm{h} v-\phi_{\mathrm{o}}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ $2 \mathrm{E}=-\phi_{\mathrm{o}}$ Subtract equation (ii) from (i), we get- $2 \mathrm{E}-\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\frac{\mathrm{hc}}{\lambda}$ $\mathrm{E}=\frac{\mathrm{hc}}{\lambda^{\prime}}-\frac{\mathrm{hc}}{\lambda}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=\mathrm{E}+\frac{\mathrm{hc}}{\lambda}$ $\frac{\mathrm{hc}}{\lambda^{\prime}}=\frac{\mathrm{E} \lambda+\mathrm{hc}}{\lambda}$ $\lambda^{\prime}=\frac{\mathrm{hc} \lambda}{\mathrm{E} \lambda+\mathrm{hc}}$
Shift-I
Dual nature of radiation and Matter
141985
Let $K_{1}$ and $K_{2}$ be the maximum kinetic energies of photo-electrons emitted when two monochromatic beams of wavelength $\lambda_{1}$ and $\lambda_{2}$, respectively are incident on a metallic surface. If $\lambda_{1}=3 \lambda_{2}$ then:
1 $K_{1}>\frac{K_{2}}{3}$
2 $\mathrm{K}_{1} \lt \frac{\mathrm{K}_{2}}{3}$
3 $\mathrm{K}_{1}=\frac{\mathrm{K}_{2}}{3}$
4 $\mathrm{K}_{2}=\frac{\mathrm{K}_{1}}{3}$
Explanation:
B Given that, $\lambda_{1}=3 \lambda_{2}$ We know that $(\mathrm{K} . \mathrm{E})_{\max }=\frac{\mathrm{hc}}{\lambda}-\phi_{\mathrm{o}}$ Case I $\mathrm{K}_{1}=\frac{\mathrm{hc}}{\lambda_{1}}-\phi_{\mathrm{o}}$ Case II $\mathrm{K}_{2}=\frac{\mathrm{hc}}{\lambda_{2}}-\phi_{\mathrm{o}}$ Then from equation (i), we get- $\mathrm{K}_{1}=\frac{\mathrm{hc}}{3 \lambda_{2}}-\phi_{\mathrm{o}}$ $3 \mathrm{~K}_{1}=\frac{\mathrm{hc}}{\lambda_{2}}-3 \phi_{\mathrm{o}}$ From equation (ii) and (iii), we get- $3 \mathrm{~K}_{1}=\mathrm{K}_{2}-2 \phi_{\mathrm{o}}$ $3 \mathrm{~K}_{1} \lt \mathrm{K}_{2}$ $\mathrm{~K}_{1} \lt \frac{\mathrm{K}_{2}}{3}$
JEE Main-28.06.2022
Dual nature of radiation and Matter
141986
A parallel beam of light of wavelength $900 \mathrm{~nm}$ and intensity $100 \mathrm{Wm}^{-2}$ is incident on a surface perpendicular to the beam. The number of photons crossing $1 \mathrm{~cm}^{2}$ area perpendicular to the beam in one second is:
1 $3 \times 10^{16}$
2 $4.5 \times 10^{16}$
3 $4.5 \times 10^{17}$
4 $4.5 \times 10^{20}$
Explanation:
B Given that, $\lambda=900 \mathrm{~nm}=9 \times 10^{-7} \mathrm{~m}$ $\mathrm{I}=100 \mathrm{~W} / \mathrm{m}^{2}$ $\mathrm{~A}=1 \mathrm{~cm}^{2}=10^{-4} \mathrm{~m}^{2}$ Number of photons in $1 \mathrm{sec}$ is given by $\mathrm{n}=\frac{\mathrm{E}_{\text {net }}}{\mathrm{E}_{\text {single photon }}}=\frac{\lambda \mathrm{AI}}{\mathrm{hc}}$ $\mathrm{n}=\frac{9 \times 10^{-7} \times 10^{-4} \times 100}{6.62 \times 10^{-34} \times 3 \times 10^{8}}$ $\mathrm{n}=4.5 \times 10^{16}$
JEE Main-26.07.2022
Dual nature of radiation and Matter
141987
Photoelectric emission from a metal begins at a frequency of $6 \times 10^{14} \mathrm{~Hz}$. The emitted electrons are fully stopped by a retarding potential of 3.3 $V$. What will be the wavelength of the incident radiation?
