If the frequency of an emitted x-ray is 1.6 x 10\(^{16}\)Hz, the accelerating potential is?
[e = 1.6 x 10\(^{-19}\)J, h = 6.63 x 10-\(^{34}\)Js]
[e = 1.6 x 10\(^{-19}\)J, h = 6.63 x 10-\(^{34}\)Js]
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Correct Answer: Option C
Explanation:
Given f = 1.6 x10\(^{16}\)H\(_{3}\), h = 6.63 x 10\(^{-39}\)Js
e = 1.6 x 10\(^{-19}\)C, V = ?
V = \(\frac{w}{e} = \frac{hf}{e}\) = \(\frac{1.6 \times 10^{16} \times 6.63 \times 10^{34}}{1.6 \times 10^{-19}}\)
= 66.3V
Given f = 1.6 x10\(^{16}\)H\(_{3}\), h = 6.63 x 10\(^{-39}\)Js
e = 1.6 x 10\(^{-19}\)C, V = ?
V = \(\frac{w}{e} = \frac{hf}{e}\) = \(\frac{1.6 \times 10^{16} \times 6.63 \times 10^{34}}{1.6 \times 10^{-19}}\)
= 66.3V