Dual Nature Of Matter Question 13
Question 13 - 01 February - Shift 2
The threshold frequency of metal is $f_0$. When the light of frequency $2 f_0$ is incident on the metal plate, the maximum velocity of photoelectron is $v_1$. When the frequency of incident radiation is increased to $5 f_0$. the maximum velocity of photoelectrons emitted is $v_2$. The ratio of $v_1$ to $v_2$ is:
(1) $\frac{v_1}{v_2}=\frac{1}{2}$
(2) $\frac{v_1}{v_2}=\frac{1}{8}$
(3) $\frac{v_1}{v_2}=\frac{1}{16}$
(4) $\frac{v_1}{v_2}=\frac{1}{4}$
Show Answer
Answer: (1)
Solution:
Formula: Photoelectric Effect
$K _{\max }=hf-hf_0$
For $f=2 f_0$
$\frac{1}{2} mV_1^{2}=2 hf f_0-hf_0=h f_0$
For $f=5 f_0$
$\frac{1}{2} m V_2{ }^{2}=5 hf_0-hf_0=4 hf_0$
$\frac{V_1}{V_2}=\frac{1}{2}$
