Question

when a certain metal was irradiated with light of frequency 4.0 ×10^16s^-1 the photoelectrons emitted had three times the kinetic energy as the kinetic energy of photoelectrons emitted when the metal was irradiated with light of frequency 2.0×10^16s^-1. calculate the critical frequency of the metal
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Answer 1

Explanation:

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Answer 2

Given:

$\begin{array}{l}K E_{1}=b\left(v_{1}-v_{0}\right) \ldots \ldots . . \text { (i) } \\K E_{2}=b\left(v_{2}-v_{0}\right) \ldots \ldots \ldots \text { (ii) }\end{array}$

$\text { Dividing equation (ii) bv7 (1), we get }$

$\frac{K E_{2}}{K E_{1}}=\frac{b\left(v_{2}-v_{0}\right)}{b\left(v_{1}-v_{0}\right)}=\frac{\left(v_{2}-v_{0}\right)}{\left(v_{1}-v_{0}\right)}$

$\begin{array}{l}\frac{K E_{2}}{K E_{1}}=3 \\\therefore 3=\frac{v_{2}-v_{0}}{v_{1}-v_{0}} \Rightarrow 3\left(v_{1}-v_{0}\right)=v_{2}-v_{0} \\\Rightarrow 3 v_{1}-v_{2}=3 v_{0}-v_{0}=2 v_{0} \\\Rightarrow 3 \times 2.0 \times 10^{16}-4 \times 10^{16}=2 v_{0} \\\Rightarrow v_{0}=\frac{2 \times 10^{16}}{2}=1 \times 10^{16} s^{-1}\end{array}$

Hence the answer is $1 \times 10^{16} s^{-1}$