Question 360: Dual Nature of Matter and Radiation

Question

Photoemission from metal surface takes place for frequencies '$$v_1$$' and '$$v_2$$' of incident rays $$\left(v_1>v_2\right)$$. Maximum kinetic energy of photoelectrons emitted is in the ratio $$1: \mathrm{K}$$. The threshold frequency of metallic surface is

Accepted Answer

$$\begin{aligned}
& (	ext { K.E. })_1=h v_1-h v_0 \quad 	ext{.....(1)}\
& (	ext { K.E. })_2=h v_2-h v_0 \quad 	ext{.....(2)}
\end{aligned}$$
Dividing Eq.(1) by Eq.(2) :
$$\frac{\left(\mathrm{K} . \mathrm{E}_1ight.}{(\mathrm{K} . \mathrm{E})_2}=\frac{1}{\mathrm{k}}=\frac{v_1-v_0}{v_2-v_0}$$
Solving for $$v_0$$ we get :
$$v_0=\frac{k v_1-v_2}{k-1}$$

Answer

$$\frac{K v_2-v_1}{K-1}$$

Answer

$$\frac{v_1-v_2}{\mathrm{~K}-1}$$

Answer

$$\frac{v_2-v_1}{K}$$

Answer

$$\frac{K v_1-v_2}{K-1}$$

Dual Nature of Matter and Radiation

Photoemission from metal surface takes place for frequencies '$v_1' and 'v_2' of incident rays \left(v_1>v_2\right). Maximum kinetic energy of photoelectrons emitted is in the ratio 1: \mathrm{K}$. The threshold frequency of metallic surface is

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