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The maximum kinetic energy of photoelectrons ejected from a photometer when it is irradiated with radiation of wavelength $400 n m$ is $1 e V$ . If the threshold energy of the metal surface is $1.9 e V$ . Select the correct statement(s).

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The maximum K.E. of photoelectrons when it is irradiated with $500 n m$ photon will be $0.42 e V$

The maximum K.E. of photoelectrons when it is irradiated with $500 n m$ photon will be $1.725 e V$

The longest wavelength which will eject photoelectrons is nearly $6100 Å$

The value of hc is $11600 eV Å$

answer is A, C, D.

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Detailed Solution

$\frac{h c}{λ} = ϕ_{0} + K_{\max} \Rightarrow \frac{h c}{4000} = (1.9 + 1) e V = 2.9 eV \Rightarrow h c = (4000) (2.9) = 11600 eV Å$

Now, for $λ = 500 nm = 5000 \overset{\circ}{A}$ , we have

$\frac{11600}{5000} = 1.9 + K_{\max}$

$\Rightarrow 2.32 = 1.9 + K_{\max}$

$\Rightarrow K_{\max} = 0.42 eV$

The longest wavelength which will eject photoelectrons is given by

$λ_{\max} = \frac{h c}{ϕ_{0}} = \frac{11600}{1.9} = 6105 \overset{\circ}{A}$

$\Rightarrow λ_{\max} = 6100 \overset{\circ}{A}$

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