NEET Solved Paper 2017 Question 38
Question: The photoelectric threshold wavelength of silver is $ 3250\times {10^{-10}}m. $ The velocity of the electron ejected from a silver surface by ultraviolet light of wavelength $ 2536\times {10^{-10}}m $ is (Given $ h=4.14\times {10^{-15}}eVs $ and $ c=3\times 180^{8}\ m{s^{-1}} $ )
Options:
A) $ \approx 0.3\times 10^{6},m{s^{-1}} $
B) $ \approx 6\times 10^{5},m{s^{-1}} $
C) $ \approx 0.6\times 10^{6},m{s^{-1}} $
D) $ \approx 61\times 10^{3},m{s^{-1}} $
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Answer:
Correct Answer: B , C
Solution:
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$ {\lambda_0}=3250\times {10^{-10}},m $
$ \Rightarrow $ $ \lambda =2536\times {10^{-10}},m $ $ \text{o }|\text{ =}\frac{1242,eV-nm}{325,nm}=3.82,$
$ \Rightarrow $ $ eV $ $ hv=\frac{1242,eV-nm}{253.6,nm}=4.89,eV $
$ \Rightarrow $ $ K{E _{\max }}=(4.89-3.82),eV=1.077,eV $
$ \Rightarrow $ $ \frac{1}{2}mv^{2}=1.077\times 1.6\times {10^{-19}} $ $ v=\sqrt{\frac{2\times 1.077\times 1.6\times {10^{-19}}}{9.1\times {10^{-31}}}} $ $ v=0.6\times 10^{6},m/s $
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$ {\lambda_0}=3250\times {10^{-10}},m $
$ \Rightarrow $ $ \lambda =2536\times {10^{-10}},m $ $ \text{o }|\text{ =}\frac{1242,eV-nm}{325,nm}=3.82,eV $ $ hv=\frac{1242,eV-nm}{253.6,nm}=4.89,eV $$ \Rightarrow $ $K{E _{\max }}=(4.89-3.82),eV=1.077,eV $
$ \Rightarrow $ $ \frac{1}{2}mv^{2}=1.077\times 1.6\times {10^{-19}} $
$ \Rightarrow $ $ v=\sqrt{\frac{2\times 1.077\times 1.6\times {10^{-19}}}{9.1\times {10^{-31}}}} $
$ \Rightarrow $ $ v=0.6\times 10^{6},m/s $
BETA
