D Solar cell change light energy into electrical energy. {|l|l|} | Photo diode | optical signal | |---|---| |LED | spontaneous emission | |Diode laser | stimulated emission | |Solar cell | ${l} { light energy into } | | { electrical energy }$ | |Photo conducting cell | photo detector | |
KERALA CEE 2012
Dual nature of radiation and Matter
142106
The photoelectric threshold wavelength for silver is $\lambda_{0}$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda\left(\lambda \lt \lambda_{0}\right)$ will be :
D According to Einstein photoelectric equationK.E $=h v-h v_{0}$ $\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}$ $= \mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)$ $= \mathrm{hc}\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{\mathrm{o}}}\right)$
Karnataka CET-2011
Dual nature of radiation and Matter
142109
There are $n_{1}$ photons of frequency $v_{1}$ in a beam of light. In an equally energetic beam there are $n_{2}$ photons of frequency $v_{2}$. Then the correct relation is:
D Two beams are equally energetic i.e. $\quad \mathrm{E}_{1}=\mathrm{E}_{2}$ $\mathrm{n}_{1} \mathrm{~h} v_{1}=\mathrm{n}_{2} \mathrm{~h} v_{2}$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{v_{2}}{v_{1}}$
Karnataka CET-2003
Dual nature of radiation and Matter
142113
The photoelectric threshold frequency of a metal is $v$. When light of frequency $4 v$ is incident on the metal, the maximum kinetic energy of the emitted photoelectron is
1 $4 \mathrm{~h} v$
2 $3 \mathrm{~h} v$
3 $5 \mathrm{~h} v$
4 $\frac{5 h v}{2}$
Explanation:
B Given that, $v_{\mathrm{o}}=v$ $\therefore \quad(\mathrm{KE})_{\max }=\mathrm{h}(4 v)-\mathrm{h} v_{\mathrm{o}}$ $(\mathrm{KE})_{\max }=4 \mathrm{~h} v-\mathrm{h} v$ $(\mathrm{KE})_{\max }=3 \mathrm{~h} v$
D Solar cell change light energy into electrical energy. {|l|l|} | Photo diode | optical signal | |---|---| |LED | spontaneous emission | |Diode laser | stimulated emission | |Solar cell | ${l} { light energy into } | | { electrical energy }$ | |Photo conducting cell | photo detector | |
KERALA CEE 2012
Dual nature of radiation and Matter
142106
The photoelectric threshold wavelength for silver is $\lambda_{0}$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda\left(\lambda \lt \lambda_{0}\right)$ will be :
D According to Einstein photoelectric equationK.E $=h v-h v_{0}$ $\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}$ $= \mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)$ $= \mathrm{hc}\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{\mathrm{o}}}\right)$
Karnataka CET-2011
Dual nature of radiation and Matter
142109
There are $n_{1}$ photons of frequency $v_{1}$ in a beam of light. In an equally energetic beam there are $n_{2}$ photons of frequency $v_{2}$. Then the correct relation is:
D Two beams are equally energetic i.e. $\quad \mathrm{E}_{1}=\mathrm{E}_{2}$ $\mathrm{n}_{1} \mathrm{~h} v_{1}=\mathrm{n}_{2} \mathrm{~h} v_{2}$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{v_{2}}{v_{1}}$
Karnataka CET-2003
Dual nature of radiation and Matter
142113
The photoelectric threshold frequency of a metal is $v$. When light of frequency $4 v$ is incident on the metal, the maximum kinetic energy of the emitted photoelectron is
1 $4 \mathrm{~h} v$
2 $3 \mathrm{~h} v$
3 $5 \mathrm{~h} v$
4 $\frac{5 h v}{2}$
Explanation:
B Given that, $v_{\mathrm{o}}=v$ $\therefore \quad(\mathrm{KE})_{\max }=\mathrm{h}(4 v)-\mathrm{h} v_{\mathrm{o}}$ $(\mathrm{KE})_{\max }=4 \mathrm{~h} v-\mathrm{h} v$ $(\mathrm{KE})_{\max }=3 \mathrm{~h} v$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142090
Identify the mismatch of the following
1 photo diode - optical signal
2 LED - spontaneous emission
3 Diode laser - stimulated emission
4 Solar cell - electrical energy into light
5 Photo conducting cell - photo detector
Explanation:
D Solar cell change light energy into electrical energy. {|l|l|} | Photo diode | optical signal | |---|---| |LED | spontaneous emission | |Diode laser | stimulated emission | |Solar cell | ${l} { light energy into } | | { electrical energy }$ | |Photo conducting cell | photo detector | |
KERALA CEE 2012
Dual nature of radiation and Matter
142106
The photoelectric threshold wavelength for silver is $\lambda_{0}$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda\left(\lambda \lt \lambda_{0}\right)$ will be :
D According to Einstein photoelectric equationK.E $=h v-h v_{0}$ $\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}$ $= \mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)$ $= \mathrm{hc}\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{\mathrm{o}}}\right)$
Karnataka CET-2011
Dual nature of radiation and Matter
142109
There are $n_{1}$ photons of frequency $v_{1}$ in a beam of light. In an equally energetic beam there are $n_{2}$ photons of frequency $v_{2}$. Then the correct relation is:
D Two beams are equally energetic i.e. $\quad \mathrm{E}_{1}=\mathrm{E}_{2}$ $\mathrm{n}_{1} \mathrm{~h} v_{1}=\mathrm{n}_{2} \mathrm{~h} v_{2}$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{v_{2}}{v_{1}}$
Karnataka CET-2003
Dual nature of radiation and Matter
142113
The photoelectric threshold frequency of a metal is $v$. When light of frequency $4 v$ is incident on the metal, the maximum kinetic energy of the emitted photoelectron is
1 $4 \mathrm{~h} v$
2 $3 \mathrm{~h} v$
3 $5 \mathrm{~h} v$
4 $\frac{5 h v}{2}$
Explanation:
B Given that, $v_{\mathrm{o}}=v$ $\therefore \quad(\mathrm{KE})_{\max }=\mathrm{h}(4 v)-\mathrm{h} v_{\mathrm{o}}$ $(\mathrm{KE})_{\max }=4 \mathrm{~h} v-\mathrm{h} v$ $(\mathrm{KE})_{\max }=3 \mathrm{~h} v$
D Solar cell change light energy into electrical energy. {|l|l|} | Photo diode | optical signal | |---|---| |LED | spontaneous emission | |Diode laser | stimulated emission | |Solar cell | ${l} { light energy into } | | { electrical energy }$ | |Photo conducting cell | photo detector | |
KERALA CEE 2012
Dual nature of radiation and Matter
142106
The photoelectric threshold wavelength for silver is $\lambda_{0}$. The energy of the electron ejected from the surface of silver by an incident wavelength $\lambda\left(\lambda \lt \lambda_{0}\right)$ will be :
D According to Einstein photoelectric equationK.E $=h v-h v_{0}$ $\frac{\mathrm{hc}}{\lambda}-\frac{\mathrm{hc}}{\lambda_{0}}$ $= \mathrm{hc}\left(\frac{1}{\lambda}-\frac{1}{\lambda_{0}}\right)$ $= \mathrm{hc}\left(\frac{\lambda_{0}-\lambda}{\lambda \lambda_{\mathrm{o}}}\right)$
Karnataka CET-2011
Dual nature of radiation and Matter
142109
There are $n_{1}$ photons of frequency $v_{1}$ in a beam of light. In an equally energetic beam there are $n_{2}$ photons of frequency $v_{2}$. Then the correct relation is:
D Two beams are equally energetic i.e. $\quad \mathrm{E}_{1}=\mathrm{E}_{2}$ $\mathrm{n}_{1} \mathrm{~h} v_{1}=\mathrm{n}_{2} \mathrm{~h} v_{2}$ $\frac{\mathrm{n}_{1}}{\mathrm{n}_{2}}=\frac{v_{2}}{v_{1}}$
Karnataka CET-2003
Dual nature of radiation and Matter
142113
The photoelectric threshold frequency of a metal is $v$. When light of frequency $4 v$ is incident on the metal, the maximum kinetic energy of the emitted photoelectron is
1 $4 \mathrm{~h} v$
2 $3 \mathrm{~h} v$
3 $5 \mathrm{~h} v$
4 $\frac{5 h v}{2}$
Explanation:
B Given that, $v_{\mathrm{o}}=v$ $\therefore \quad(\mathrm{KE})_{\max }=\mathrm{h}(4 v)-\mathrm{h} v_{\mathrm{o}}$ $(\mathrm{KE})_{\max }=4 \mathrm{~h} v-\mathrm{h} v$ $(\mathrm{KE})_{\max }=3 \mathrm{~h} v$