Light of frequency $8 \times 10^{15} H z$ is incident on a substance of photoelectric work function 6.125 eV. The maximum kinetic energy of the emitted photoelectrons is

17 eV

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22 eV

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27 eV

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37 eV

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Solution

The correct option is C
27 eV

$E = h v = 6.6 \times 10^{- 34} \times 8 \times 10^{15} = 5.28 \times 10^{- 18} J = 33 e V B y u s i n g E = W_{0} + K_{m a x} \Rightarrow K_{m a x} = E - W_{0}$
$= 33 - 6.125 = 27 e V$

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Light of frequency 8×1015Hz is incident on a substance of photoelectric work function 6.125 eV. The maximum kinetic energy of the emitted photoelectrons is

The correct option is C 27 eV E=hv=6.6×10−34×8×1015=5.28×10−18J=33eV By using E=W0+Kmax⇒Kmax=E−W0 =33−6.125=27 eV

Light of frequency $8 \times 10^{15} H z$ is incident on a substance of photoelectric work function 6.125 eV. The maximum kinetic energy of the emitted photoelectrons is

The correct option is C
27 eV

$E = h v = 6.6 \times 10^{- 34} \times 8 \times 10^{15} = 5.28 \times 10^{- 18} J = 33 e V B y u s i n g E = W_{0} + K_{m a x} \Rightarrow K_{m a x} = E - W_{0}$
$= 33 - 6.125 = 27 e V$