Em Waves Question 85
Question: In the experiment on photoelectric effect, the graph between $ E _{Kmax} $ is found to be a straight line as shown in figure. The threshold frequency and Planck’s constant according to this graph are
Options:
A) $ 3.33\times 10^{18}{{s}^{-1}},6\times {{10}^{-34}}J\text{-}s $
B) $ 6\times 10^{18}{{s}^{-1}},6\times {{10}^{-34}}J\text{-}s $
C) $ 2.66\times 10^{18}{{s}^{-1}},4\times {{10}^{-34}}J\text{-}s $
D) $ 4\times 10^{18}{{s}^{-1}},3\times {{10}^{-34}}J\text{-}s $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] $ {K _{\max }}=hv-W $
$ \omega $ is the intercept on y-axis and h is the slope.
$ \
Therefore h=\frac{2.4\times {{10}^{-15}}}{4\times 10^{18}}=6\times {{10}^{-34}}Js $
$ W=2\times {{10}^{-15}}J $
$ \Rightarrow hv _{0}=2\times {{10}^{-15}} $ Or $ v _{0}=3.33\times 10^{18}s $
BETA
