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JEE PHYSICS MCQ: ATOMS

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Application +4 | -1
X-ray of wavelength λ=2A∘\lambda = 2\mathop A\limits^ \circ is emitted from the metal target. The potential difference applied across the cathode and the metal target is
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Correct option: C
6200 V
Option C (6200 V) is correct because it is the potential difference required to produce X-rays of the given wavelength, calculated using V=hceΞ»V = \frac{{hc}}{{e\lambda}}.

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