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A photosensitive metallic surface has work function $\phi$ . If photon of energy $3 \phi$ fall on this surface, the electron comes out with a maximum velocity of $6 \times 10^6 \mathrm{~ms}^{-1}$ . When the photon energy is increased to $9 \phi$ , there maximum velocity of photoelectron will be:

Option: 1
$12 \times 10^6 \mathrm{~m} / \mathrm{s}$

Option: 2
$6 \times 10^6 \mathrm{~m} / \mathrm{s}$

Option: 3
$3 \times 10^6 \mathrm{~m} / \mathrm{s}$

Option: 4
$24 \times 10^6 \mathrm{~m} / \mathrm{s}$

Answers (1)

best_answer

The relation between the energy and the work function of the electron is given as:

$\mathrm{E}=\phi+\frac{1}{2} \mathrm{mv}^2$

For the total energy of $3 \phi$ , the expression can be written as:

$\begin{aligned} & 3 \phi=\phi+\frac{1}{2} \mathrm{mv}_1^2 \\ & \frac{1}{4} \mathrm{m} v_1^2=\phi \end{aligned}$

Now, the energy provided is $9 \phi$ . So, it can be written as:

$\begin{aligned} & 9 \phi=\phi+\frac{1}{2} \mathrm{mv}_2^2 \\ & 8 \phi=\frac{1}{2} \mathrm{mv}_2^2 \end{aligned}$

Substitute the value of $\phi$ .

$\begin{aligned} & 8 \times \frac{1}{4} \mathrm{mv}_1^2=\frac{1}{2} \mathrm{mv}_2^2 \\ & \mathrm{v}_2=\sqrt{4 \mathrm{v}_1^2} \\ & \Rightarrow \sqrt{4 \times 36 \times 10^{12}} \\ & \mathrm{v}_2=12 \times 10^6 \mathrm{~m} / \mathrm{s} \end{aligned}$

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