JEE Main 2017 · 2 Apr · Shift 1 · Physics Q21

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PhysicsSingle correct+4 / -1Question 21 of 90

An electron beam is accelerated by a potential difference V to hit a metallic target to produce X-rays. It produces continuous as well as characteristic X-rays. If $\lambda_{min}$ is the smallest possible wavelength of X-ray in the spectrum, the variation of $\log\lambda_{min}$ with log V is correctly represented in :

A straight line falling with log V

Graph of log λ_min against log V: a straight line falling to the right.

A straight line through the origin rising with log V

Graph of log λ_min against log V: a straight line rising from the origin.

A straight line rising with log V from a positive intercept

Graph of log λ_min against log V: a straight line rising gently from a positive intercept.

A horizontal line

Answer and worked solution

Official answer: A

$\lambda_{min}=\dfrac{hc}{eV}$ , so $\log\lambda_{min}=\log\dfrac{hc}e-\log V$ : a straight line of slope −1, falling. That is graph (1).

Topics:X-rays

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JEE Main 2017 · 2 Apr · Shift 1, Physics question 21