Light of wavelength 5000 $\circ A$ falls on a sensitive plate with photoelectric work function of 1.9 eV. The kinetic energy of the photoelectron emitted will be

0.58 eV

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2.48 eV

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1.24 eV

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1.16 eV

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Solution

The correct option is A
0.58 eV

$E = W_{0} + K_{m a x}; E = \frac{12375}{5000} = 2.475 e V ∴ K_{m a x} = E - W_{0} = 2.475 - 1.9 = 0.57 e V$

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[CBSE PMT 1998]

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Light of wavelength 5000 ∘A falls on a sensitive plate with photoelectric work function of 1.9 eV. The kinetic energy of the photoelectron emitted will be

The correct option is A 0.58 eV E=W0+Kmax; E=123755000=2.475 eV ∴Kmax=E−W0=2.475−1.9=0.57eV

Light of wavelength 5000 $\circ A$ falls on a sensitive plate with photoelectric work function of 1.9 eV. The kinetic energy of the photoelectron emitted will be

The correct option is A
0.58 eV

$E = W_{0} + K_{m a x}; E = \frac{12375}{5000} = 2.475 e V ∴ K_{m a x} = E - W_{0} = 2.475 - 1.9 = 0.57 e V$