142347
Assertion: In photoemissive cell inert gas is used. Reason: Inert gas in the photoemissive cell gives greater current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A The photoemissive cell contain two electrodes are enclosed in a glass bulb which may be evacuated or contain an inert gas at low pressure. An inert gas in the cell gives greater current but causes a time lag in the response of the cell to very rapid changes of radiation which may make it unsuitable for some purpose.
AIIMS-2010
Dual nature of radiation and Matter
142353
The momentum of a proton is $2.5 \times 10^{-29} \mathrm{~kg}-$ $\mathrm{m} / \mathbf{s}$. Its frequency will be
142354
Ultraviolet radiations of $6.2 \mathrm{eV}$ fall on an aluminium surface (work function $4.2 \mathrm{eV}$ ). The kinetic energy in joules of the fastest electron emitted is approximately
1 $3.2 \times 10^{-21} \mathrm{~J}$
2 $3.2 \times 10^{-19} \mathrm{~J}$
3 $3.2 \times 10^{-17} \mathrm{~J}$
4 $3.2 \times 10^{-15} \mathrm{~J}$
Explanation:
B Given that, $\mathrm{E}=\mathrm{hv}=6.2 \mathrm{eV}$, $\phi_{0}=4.2 \mathrm{eV}$ We know that, by photo electric effect $\mathrm{KE}=\mathrm{h} v-\phi_{0}$ $\mathrm{KE}=(6.2-4.2) \mathrm{eV}$ $\mathrm{KE}=2 \mathrm{eV}$ $\mathrm{KE}=2 \times 1.6 \times 10^{-19} \mathrm{~J}$ $\mathrm{KE}=3.2 \times 10^{-19} \mathrm{~J}$
SRMJEEE - 2010
Dual nature of radiation and Matter
142339
For a photocell, work function is ' $W_{0}$ ' and stopping potential is ' $V$ '. The wavelength of the incident radiation is $(h=$ Plank's constant, $C=$ velocity of light, $\mathrm{e}=$ charge of electron)
D We know that, The relation in photo electric effect, $\mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}_{0}$ $\frac{\mathrm{hc}}{\lambda}=\mathrm{W}_{0}+\mathrm{eV}$ $\lambda=\frac{\mathrm{hc}}{\mathrm{W}_{0}+\mathrm{eV}}$
NEET Test Series from KOTA - 10 Papers In MS WORD
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Dual nature of radiation and Matter
142347
Assertion: In photoemissive cell inert gas is used. Reason: Inert gas in the photoemissive cell gives greater current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A The photoemissive cell contain two electrodes are enclosed in a glass bulb which may be evacuated or contain an inert gas at low pressure. An inert gas in the cell gives greater current but causes a time lag in the response of the cell to very rapid changes of radiation which may make it unsuitable for some purpose.
AIIMS-2010
Dual nature of radiation and Matter
142353
The momentum of a proton is $2.5 \times 10^{-29} \mathrm{~kg}-$ $\mathrm{m} / \mathbf{s}$. Its frequency will be
142354
Ultraviolet radiations of $6.2 \mathrm{eV}$ fall on an aluminium surface (work function $4.2 \mathrm{eV}$ ). The kinetic energy in joules of the fastest electron emitted is approximately
1 $3.2 \times 10^{-21} \mathrm{~J}$
2 $3.2 \times 10^{-19} \mathrm{~J}$
3 $3.2 \times 10^{-17} \mathrm{~J}$
4 $3.2 \times 10^{-15} \mathrm{~J}$
Explanation:
B Given that, $\mathrm{E}=\mathrm{hv}=6.2 \mathrm{eV}$, $\phi_{0}=4.2 \mathrm{eV}$ We know that, by photo electric effect $\mathrm{KE}=\mathrm{h} v-\phi_{0}$ $\mathrm{KE}=(6.2-4.2) \mathrm{eV}$ $\mathrm{KE}=2 \mathrm{eV}$ $\mathrm{KE}=2 \times 1.6 \times 10^{-19} \mathrm{~J}$ $\mathrm{KE}=3.2 \times 10^{-19} \mathrm{~J}$
SRMJEEE - 2010
Dual nature of radiation and Matter
142339
For a photocell, work function is ' $W_{0}$ ' and stopping potential is ' $V$ '. The wavelength of the incident radiation is $(h=$ Plank's constant, $C=$ velocity of light, $\mathrm{e}=$ charge of electron)
D We know that, The relation in photo electric effect, $\mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}_{0}$ $\frac{\mathrm{hc}}{\lambda}=\mathrm{W}_{0}+\mathrm{eV}$ $\lambda=\frac{\mathrm{hc}}{\mathrm{W}_{0}+\mathrm{eV}}$
142347
Assertion: In photoemissive cell inert gas is used. Reason: Inert gas in the photoemissive cell gives greater current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A The photoemissive cell contain two electrodes are enclosed in a glass bulb which may be evacuated or contain an inert gas at low pressure. An inert gas in the cell gives greater current but causes a time lag in the response of the cell to very rapid changes of radiation which may make it unsuitable for some purpose.
AIIMS-2010
Dual nature of radiation and Matter
142353
The momentum of a proton is $2.5 \times 10^{-29} \mathrm{~kg}-$ $\mathrm{m} / \mathbf{s}$. Its frequency will be
142354
Ultraviolet radiations of $6.2 \mathrm{eV}$ fall on an aluminium surface (work function $4.2 \mathrm{eV}$ ). The kinetic energy in joules of the fastest electron emitted is approximately
1 $3.2 \times 10^{-21} \mathrm{~J}$
2 $3.2 \times 10^{-19} \mathrm{~J}$
3 $3.2 \times 10^{-17} \mathrm{~J}$
4 $3.2 \times 10^{-15} \mathrm{~J}$
Explanation:
B Given that, $\mathrm{E}=\mathrm{hv}=6.2 \mathrm{eV}$, $\phi_{0}=4.2 \mathrm{eV}$ We know that, by photo electric effect $\mathrm{KE}=\mathrm{h} v-\phi_{0}$ $\mathrm{KE}=(6.2-4.2) \mathrm{eV}$ $\mathrm{KE}=2 \mathrm{eV}$ $\mathrm{KE}=2 \times 1.6 \times 10^{-19} \mathrm{~J}$ $\mathrm{KE}=3.2 \times 10^{-19} \mathrm{~J}$
SRMJEEE - 2010
Dual nature of radiation and Matter
142339
For a photocell, work function is ' $W_{0}$ ' and stopping potential is ' $V$ '. The wavelength of the incident radiation is $(h=$ Plank's constant, $C=$ velocity of light, $\mathrm{e}=$ charge of electron)
D We know that, The relation in photo electric effect, $\mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}_{0}$ $\frac{\mathrm{hc}}{\lambda}=\mathrm{W}_{0}+\mathrm{eV}$ $\lambda=\frac{\mathrm{hc}}{\mathrm{W}_{0}+\mathrm{eV}}$
142347
Assertion: In photoemissive cell inert gas is used. Reason: Inert gas in the photoemissive cell gives greater current.
1 If both Assertion and Reason are correct and reason is the correct explanation of Assertion.
2 If both Assertion and Reason are correct, but Reason is not the correct explanation of Assertion
3 If Assertion is correct but Reason is incorrect
4 If both the Assertion and Reason are incorrect
Explanation:
A The photoemissive cell contain two electrodes are enclosed in a glass bulb which may be evacuated or contain an inert gas at low pressure. An inert gas in the cell gives greater current but causes a time lag in the response of the cell to very rapid changes of radiation which may make it unsuitable for some purpose.
AIIMS-2010
Dual nature of radiation and Matter
142353
The momentum of a proton is $2.5 \times 10^{-29} \mathrm{~kg}-$ $\mathrm{m} / \mathbf{s}$. Its frequency will be
142354
Ultraviolet radiations of $6.2 \mathrm{eV}$ fall on an aluminium surface (work function $4.2 \mathrm{eV}$ ). The kinetic energy in joules of the fastest electron emitted is approximately
1 $3.2 \times 10^{-21} \mathrm{~J}$
2 $3.2 \times 10^{-19} \mathrm{~J}$
3 $3.2 \times 10^{-17} \mathrm{~J}$
4 $3.2 \times 10^{-15} \mathrm{~J}$
Explanation:
B Given that, $\mathrm{E}=\mathrm{hv}=6.2 \mathrm{eV}$, $\phi_{0}=4.2 \mathrm{eV}$ We know that, by photo electric effect $\mathrm{KE}=\mathrm{h} v-\phi_{0}$ $\mathrm{KE}=(6.2-4.2) \mathrm{eV}$ $\mathrm{KE}=2 \mathrm{eV}$ $\mathrm{KE}=2 \times 1.6 \times 10^{-19} \mathrm{~J}$ $\mathrm{KE}=3.2 \times 10^{-19} \mathrm{~J}$
SRMJEEE - 2010
Dual nature of radiation and Matter
142339
For a photocell, work function is ' $W_{0}$ ' and stopping potential is ' $V$ '. The wavelength of the incident radiation is $(h=$ Plank's constant, $C=$ velocity of light, $\mathrm{e}=$ charge of electron)
D We know that, The relation in photo electric effect, $\mathrm{eV}=\frac{\mathrm{hc}}{\lambda}-\mathrm{W}_{0}$ $\frac{\mathrm{hc}}{\lambda}=\mathrm{W}_{0}+\mathrm{eV}$ $\lambda=\frac{\mathrm{hc}}{\mathrm{W}_{0}+\mathrm{eV}}$