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Previous Years AIEEE/JEE Mains Questions
Q. The surface of a metal is illuminated with the light of 400 nm. The kinetic energy of the ejected photoelectrons was found to be 1.68 eV. The work function of the metal is : (hc = 1240 eV-nm)
(1) 1.51 eV (2) 1.68 eV (3) 3.09 eV (4) 1.41 eV
[AIEEE - 2009]
Ans. (4)
$\mathrm{E}_{\mathrm{k}}=\frac{\mathrm{hc}}{\lambda}-\phi_{0} \Rightarrow 1.68=\frac{12400}{4000}-\phi_{0}$
By solving it $\phi_{0}=1.42 \mathrm{eV}$
Q. Statement-1 : When ultraviolet light is incident on a photocell, its stopping potential is V0 and the maximum kinetic energy of the photoelectrons is $\mathrm{K}_{\mathrm{max}}$. When the ultraviolet light is replaced by X-rays, both $\mathrm{V}_{0}$ and $\mathrm{K}_{\mathrm{max}}$ increase.
Statement-2 : Photoelectrons are emitted with speeds ranging from zero to a maximum value because of the range of frequencies present in the incident light
(1) Statement–1 is true, Statement–2 is false
(2) Statement–1 is true, Statement–2 is true; Statement–2 is the correct explanation of Statement– 1
(3) Statement–1 is true, Statement–2 is true; Statement–2 is not the correct explanation of Statement– 1
(4) Statement–1 is false, Statement–2 is true
[AIEEE - 2010]
Ans. (1)
Speed of emitted electrons is independent of frequency of incident light.
Q. This question has Statememtn-1 and Statement-2. Of the four choices given after the statements, choose the one that best describes the two statements.
Statement–1 : A metallic surface is irradiated by a monochromatic light of frequency $\mathrm{v}>\mathrm{v}_{0}$ (the threshold frequency). The maximum kinetic energy and the stopping potential are $\mathrm{K}_{\max }$ and $\mathrm{V}_{0}$ respectively. If the frequency incident on the surface is doubled, both the $\mathrm{K}_{\max }$ and $\mathrm{V}_{0}$ are also boubled.
Statement-2 : The maximum kinetic energy and the stopping potential of photoelectrons emitted from a surface are linearly dependent on the frequency of incident light.
(1) Statement–1 is true, Statement–2 is true, Statement–2 is not the correct explanationof Statement– 1
(2) Statement–1 is false, Statement–2 is true
(3) Statement–1 is true, Statement–2 is false
(4) Statement–1 is true, Statement–2 is true, Statement–2 is the correct explanation of Statement– 1
[AIEEE-2011]
Ans. (2)
Q. The anode voltage of photocell is kept fixed. The wavelength $\lambda$ of the light falling on the cathode is gradually changed. The plate current I of the photocell varies as follows
[AIEEE-2013]
[AIEEE-2013]
Ans. (4)
For constant intensity as wavelength decreases energy of photons increases and number of photons decreases. So it may seem that current should decrease. But the probability that a photon will be successful in emitting an electron will also increase. So as wavelength decreases current increases
Q. Radiation of wavelength $\lambda$, is incident on a photocell. The fastest emitted electron has speed v. If the wavelength of changed to $\frac{3 \lambda}{4}$, the speed of the fastest emitted electron will be :-
[JEE Main-2016]
[JEE Main-2016]
Ans. (2)
$\mathrm{E}=(\mathrm{KE})_{\max }+\mathrm{f}$
$\left[\frac{\mathrm{hc}}{\lambda}=(\mathrm{KE})_{\max }+\phi\right] \ldots .$ (1)
\frac{4}{3} \frac{\mathrm{hc}}{\lambda}=\left(\frac{4}{3} \mathrm{KE}_{\max }+\frac{\phi}{3}\right)+\phi
Comments
Rampratap
July 5, 2021, 4:16 p.m.
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NONE OF YOUR BUISNESS
May 16, 2021, 8:44 p.m.
WTF????????????????????
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Priyanshu tiwari
April 21, 2021, 1:59 p.m.
Statement type question are tougher than numerical type question
TANISHKA
April 15, 2021, 3:27 a.m.
JUST STARTED WITH THE TOPIC...THE QUESTIONS SEEM QUITE EASY...
BUT THANKYOU SO MUCH !