Light of wavelength 200 $n m$ falls on an aluminium plate. If work function of plate is 2 $e V$ , then kinetic energy of the fastest electron could be

4. 21

e V

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e V

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Solution

The correct option is A 4. 21 $e V$
Energy of a photon, $E_{p h o t o n} = \frac{1242}{λ} = \frac{1242}{200} = 6.21 e V$

Maximum kinetic energy of an electron, $K E_{m a x} = E_{p h o t o n} - ϕ = 6.21 - 2 = 4.21 e V$

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Light of wavelength 200 nm falls on an aluminium plate. If work function of plate is 2 eV, then kinetic energy of the fastest electron could be

The correct option is A 4. 21 eVEnergy of a photon, Ephoton=1242λ=1242200=6.21 eV Maximum kinetic energy of an electron, KEmax=Ephoton−ϕ=6.21−2=4.21 eV

Light of wavelength 200 $n m$ falls on an aluminium plate. If work function of plate is 2 $e V$ , then kinetic energy of the fastest electron could be

The correct option is A 4. 21 $e V$
Energy of a photon, $E_{p h o t o n} = \frac{1242}{λ} = \frac{1242}{200} = 6.21 e V$

Maximum kinetic energy of an electron, $K E_{m a x} = E_{p h o t o n} - ϕ = 6.21 - 2 = 4.21 e V$