SATHEE: Atomic Structure Question 427

Atomic Structure Question 427

Question: The threshold frequency of a metal is $1 \times 10^{15} s^{- 1} .$ The ratio of the maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies $1.5 \times 10^{15} s^{- 1}$ and $2.0 \times 10^{15} s^{- 1}$ respectively would be

Options:

A) 4 : 3

B) 1 : 2

C) 2 : 1

D) 3 : 4

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Answer:

Correct Answer: B

Solution:

$\frac{{(K E)}_{1}}{{(K E)}_{2}} = \frac{v_{1} - v_{0}}{v_{2} - v_{0}}$

$= \frac{(1.5 \times 10^{15} - 1 \times 10^{15})}{(2.0 \times 10^{15} - 1 \times 10^{15})} = \frac{1}{2}$

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Atomic Structure Question 427

Question: The threshold frequency of a metal is 1×1015s−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

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Question: The threshold frequency of a metal is $1 \times 10^{15} s^{- 1} .$ The ratio of the maximum kinetic energies of the photoelectrons when the metal is irradiated with radiations of frequencies $1.5 \times 10^{15} s^{- 1}$ and $2.0 \times 10^{15} s^{- 1}$ respectively would be