Dual Nature Of Light Question 395
Light of wavelength $ \text{200 }\overset{o}{\mathop{A}}, $ falls on aluminum surface. Work function of aluminum is 4.2 eV. What is the kinetic energy of the fastest emitted photoelectrons?
Options:
2eV
1eV
4eV
D) 0.2eV
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Answer:
Correct Answer: A
Solution:
[a] By Einstein’s equation of photo-electric effect, the maximum kinetic energy of emitted photo-electrons is given by $ E_{k}=hv-W\text{ or }E_{k}=\frac{hc}{\lambda }-W $ Where, h = Planck’s constant v = frequency of incident light W = work function of metal $ \lambda $ = wavelength of incident light $ E_{k}=\frac{6.6\times {{10}^{-,34}}\times 3\times 10^{8}}{2000\times {{10}^{-,10}}\times 1.6\times {{10}^{-19}}}eV=4.2eV $ So, $ E_{k}=2eV $
BETA
