The light of wavelength $5000 \circ A$ falls on a photosensitive plate of work function $1.9 e V$ . The kinetic energy of the emitted photoelectrons is :

0.58 e V

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2.48 e V

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1.24 e V

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1.18 e V

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Solution

The correct option is A $0.58 e V$
The energy of photon $E = \frac{h c}{λ}$
$= \frac{12375}{λ} e V$
$E = \frac{12375}{5000}$
$= 2.48 e V$
The kinetic energy of equation of photoelectric
$E_{k} = E - W$
$= 0.58 e V$ .

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The light of wavelength 5000∘A falls on a photosensitive plate of work function 1.9 eV. The kinetic energy of the emitted photoelectrons is :

The correct option is A 0.58 eVThe energy of photon E=hcλ=12375λeVE=123755000=2.48 eVThe kinetic energy of equation of photoelectricEk=E−W=0.58 eV.

The light of wavelength $5000 \circ A$ falls on a photosensitive plate of work function $1.9 e V$ . The kinetic energy of the emitted photoelectrons is :

The correct option is A $0.58 e V$
The energy of photon $E = \frac{h c}{λ}$
$= \frac{12375}{λ} e V$
$E = \frac{12375}{5000}$
$= 2.48 e V$
The kinetic energy of equation of photoelectric
$E_{k} = E - W$
$= 0.58 e V$ .