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Surface of certain metal is first illuminated with light of wavelength $\lambda _{1}=350\: nm$ and then, by light of wavelength $\lambda _{2}=540\: nm$ . It is found that the maximum speed of the photo electrons in the two cases differ by a factor of 2. The work function of the metal (in eV) is close to :

( Energy of photon = $\frac{1240}{\lambda (in\: nm)}eV$ )
Option 1)
$1.8\: \: \:$
Option 2)
$2.5\: \: \:$
Option 3)
$5.6\: \:$
Option 4)
$1.4$

Answers (1)

best_answer

Conservation of energy -

$h\nu = \phi _{0}+\frac{1}{2}mv^{2}_{max}$

$h\nu = h\nu _{0}+\frac{1}{2}mv^{2}_{max}$

$h\left ( \nu-\nu _{0} \right )=\frac{1}{2}mv^{2}_{max}$

$where, h - Plank's constant\ \nu - Frequency\ \nu_{0} - threshold\ frequency\ \phi_{0}- work function$

- wherein

Given the maximum speed of photo electron differ by a factor of 2

$\Rightarrow \frac{V_{1}}{V_{2}} = 2$

$\therefore \frac{hc}{\lambda _{1}} = \phi + \frac{1}{2} m(2v) ^{2} - (1)$

$\therefore \frac{hc}{\lambda _{2}} = \phi + \frac{1}{2} mv ^{2} - (2)$

from (2)

$\frac{1}{2} mv ^{2} = \frac{hc}{\lambda _{2}} - \phi \ \ \ - (3)$

from (3) and (1)

$\frac{hc}{\lambda _{1}} = \phi + 4 \left ( \frac{hc}{\lambda _{2}} - \phi \right )$

$\phi = \frac{hc}{3} \left ( \frac{4}{\lambda _{2}} - \frac{1}{\lambda _{1}} \right )$

$= \frac{1240}{3} \left ( \frac{4}{540} - \frac{1}{350} \right )$

= 1.8 eV

Option 1)
$1.8\: \: \:$
Option 2)

$2.5\: \: \:$

Option 3)

$5.6\: \:$

Option 4)

$1.4$

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