Photoelectric emission from a metal begins at a frequency of $6 \times 10^{14} H z .$ The emitted electrons are fully stopped by a retarding potential of 3.3V. Find the wavelength (in nm) of the incident radiation. Take $h = 6.6 \times 10^{- 34} J s .$ $e = 1.6 \times 10^{- 19} C$

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answer is 214.

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Detailed Solution

$e V_{0} = h (v - v_{0}) \Rightarrow (1.6 \times 10^{- 19}) \times 3.3 = (6.6 \times 10^{- 34}) \times (v - 6 \times 10^{14}) \Rightarrow v = 1.4 \times 10^{15} H z$

$∴ λ = \frac{c}{v} = \frac{3 \times 10^{8}}{1.4 \times 10^{15}} = 2.14 \times 10^{- 7} m = 214 n m$

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