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*Simulator*

**Previous Years AIEEE/JEE Mains Questions**

(1) 1.51 eV (2) 1.68 eV (3) 3.09 eV (4) 1.41 eV

**[AIEEE – 2009]**

**Sol.**(4)

$\mathrm{E}_{\mathrm{k}}=\frac{\mathrm{hc}}{\lambda}-\phi_{0} \Rightarrow 1.68=\frac{12400}{4000}-\phi_{0}$

By solving it $\phi_{0}=1.42 \mathrm{eV}$

**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]**

**Sol.**(1)

Speed of emitted electrons is independent of frequency of incident light.

**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]**

**Sol.**(2)

** [AIEEE-2013]**

**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

**[JEE Main-2016]**

**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