NEET-XII-Physics
42: Photoelectric Effect and Wave Particle Duality
- #9If the frequency of light in a photoelectric experiment is doubled, the stopping potential will
(a) be doubled
(b) be halved
(c) become more than double
(d) become less than doubledigAnsr: cAns : (c) become more than double
According to Einstein's equation of photoelectric effect,
`` e{V}_{0}=hv-\phi ``
`` \Rightarrow {V}_{0}=\frac{hv-\phi }{e}...\left(1\right)``
Here, V0 = stopping potential
v = frequency of light
`` \phi `` = work function
Let the new frequency of light be 2ν and the corresponding stopping potential be V0'.
Therefore,
`` e{V}_{0}\text{'}=2hv-\phi ``
`` {V}_{0}\text{'}=\frac{2hv-\phi }{e}...\left(2\right)``
Multiplying both sides of equation (1) by 2, we get:
`` 2{V}_{0}=\frac{2hv-2\phi }{e}...\left(3\right)``
Now if we compare (2) and (3), it can be observed that:
`` \frac{2hv-\phi }{e}>\frac{2hv-2\phi }{e}``
`` \Rightarrow {V}_{0}\text{'}>2{V}_{0}``
It is clear from the above equation that if the frequency of light in a photoelectric experiment is doubled, the stopping potential will be more than doubled.
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