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19/08/2021 12:48 pm
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The threshold frequency of metal is 1 × 1015 s-1. The ratio of the maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies 1.5 × 1015s-1 and 2.0 × 1015s-1 respectively would be
(a) 4 : 3
(b) 1 : 2
(c) 2 : 1
(d) 3 : 4
1 Answer
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19/08/2021 12:52 pm
Correct answer: (b) 1 : 2
Explanation:
\(\frac{(KE)_1}{(KE)_2}\) = \(\frac{v_1 - v_0}{v_2 - v_0}\)
= \(\frac{(1.5 \times 10^{15} - 1 \times 10^{15})}{(2.0 \times 10^{15} - 1 \times 10^{15})}\)
= \(\frac{1}{2}\) = 1 : 2