Photoelectric Effect – JEE Main Previous Year Questions with Solutions

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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]

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Sol. (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]

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Sol. (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]

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Sol. (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]

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Sol. (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]

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Sol. (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

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Comments
  • April 21, 2021 at 1:59 pm

    Statement type question are tougher than numerical type question

    0
  • April 15, 2021 at 3:27 am

    JUST STARTED WITH THE TOPIC…THE QUESTIONS SEEM QUITE EASY…
    BUT THANKYOU SO MUCH !

    2
  • December 26, 2020 at 4:06 pm

    Easyiest question ever😅😅

    1
  • October 19, 2020 at 8:02 pm

    Nice start to encourage jee aspirants with previous year questions

    0
  • WQW
    September 8, 2020 at 6:23 pm

    PRISTIGIOUS QUESTION

    0
  • April 21, 2020 at 2:54 pm

    Something error

    0
  • March 25, 2020 at 9:28 pm

    Fantastic

    0