JEE Advanced 2020 Full Test 5 Paper 2

Question : 10 of 54

Marks: +1, -0

In a photoelectric effect experiment, the frequency of incident radiation is increased n times (n>1). Then

Saturation current remains constant
Stopping potential becomes more than n times of initial stopping potential
The difference of the two stopping potentials is equal to (n−1)(Vs+‌
φ
e
), where f is the work function and Vs is the initial stopping potential
The maximum kinetic energy of the photoelectrons increases.

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