The threshold frequency for a metallic surface corresponds to an energy of $6.2 e V$ and the stopping potential for a radiation incident on this surface is $5 V$ . The incident radiation lies in:

Ultra-Violet region

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Infrared region

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Visible region

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X-ray region

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Solution

The correct option is C Ultra-Violet region
Stopping potential $= (i n c i d e n t e n e r g y - w o r k f u n c t i o n) / e$
5 ev $=$ incident energy - work function
5ev $=$ incident energy - 6.2ev
( $∵$ work function corresponds to energy of these hold frequency)
$∴ i n c i d e n t e n e r g y = (5 + 6.2) e v = 11.2 e v$

Energy $= \frac{h c}{λ}$

$11.2 e v = \frac{1240}{λ} (h c = 1240 e v - n m)$

$λ = \frac{1240}{11.2} n m$

$= 110 n m$
$= 1100 A$
So, radiation is ultra-violet region rays, because it has wavelength of $100 A^{0} t o 4000 A^{0}$ .

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The threshold frequency for a metallic surface corresponds to an energy of 6.2 eV and the stopping potential for a radiation incident on this surface is 5 V. The incident radiation lies in:

The threshold frequency for a metallic surface corresponds to an energy of $6.2 e V$ and the stopping potential for a radiation incident on this surface is $5 V$ . The incident radiation lies in: