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5 volt of stopping potential is needed for the photo electrons emitted out of a surface of work function 2.2 eV by the radiation of wavelength

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$861 \overset{0}{A}$

$3000 \overset{0}{A}$

$3444 \overset{0}{A}$

$1722 \overset{0}{A}$

answer is A.

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Detailed Solution

${(KE)}_{\max}$ of ejected photo electron $= {eV}_{0} = 5 eV$

$∴$ Energy of incident photon $= (5 + 2 .2) eV = 7 .2 eV$

$∴ \frac{h c}{λ} = 7 .2 eV \Rightarrow λ = \frac{h c}{7 .2} = \frac{12400}{7 .2} \overset{0}{A} = 1722 \overset{0}{A}$

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