JEE Main 2016 · 3 Apr · Shift 1 · Physics Q21

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PhysicsSingle correct+4 / -1Question 21 of 90

Radiation of wavelength λ, is incident on a photocell. The fastest emitted electron has speed $v$ . If the wavelength is changed to $\dfrac{3\lambda}4$ , the speed of the fastest emitted electron will be :

>v\left(\dfrac43\right)^{1/2}

<v\left(\dfrac43\right)^{1/2}

=v\left(\dfrac43\right)^{1/2}

=v\left(\dfrac34\right)^{1/2}

Answer and worked solution

Official answer: A

$\frac12mv^2=\frac{hc}\lambda-\phi$ , and with $\frac{3\lambda}4$ :
$\tfrac12mv'^2=\tfrac43\frac{hc}\lambda-\phi=\tfrac43\left(\tfrac12mv^2\right)+\tfrac13\phi>\tfrac43\left(\tfrac12mv^2\right).$
So $v'>v\left(\frac43\right)^{1/2}$ .

Topics:Photoelectric effect

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JEE Main 2016 · 3 Apr · Shift 1, Physics question 21