Light of 500nm is incident on a metal with work function 2.28ev .The de broglie wavelength of the emitted electron is 1) less than 2.8 ×10*-12m 2)less than 2.8×10*-10m 3) less than 2.8×10*-9m 4) greater than 2.8 ×10*9m

Question

Physics Expert · Answer

Calculate the energy of the photon.

Calculate the maximum kinetic energy as per the Einstein photoelectric emmission.

Calculate the debroglie wavelength by the equation as below.

Hence, the correct option is B.

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Light of 500nm is incident on a metal with work function 2.28ev .The de broglie wavelength of the emitted electron is 1) less than 2.8 ×10-12m 2)less than 2.8×10-10m 3) less than 2.8×10-9m 4) greater than 2.8 ×109m

Light of 500nm is incident on a metal with work function 2.28ev .The de broglie wavelength of the emitted electron is
1) less than 2.8 ×10-12m

2)less than 2.8×10-10m
3) less than 2.8×10-9m
4) greater than 2.8 ×109m

1 Answers

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Light of 500nm is incident on a metal with work function 2.28ev .The de broglie wavelength of the emitted electron is 1) less than 2.8 ×10*-12m 2)less than 2.8×10*-10m 3) less than 2.8×10*-9m 4) greater than 2.8 ×10*9m

Light of 500nm is incident on a metal with work function 2.28ev .The de broglie wavelength of the emitted electron is1) less than 2.8 ×10*-12m 2)less than 2.8×10*-10m3) less than 2.8×10*-9m4) greater than 2.8 ×10*9m

1 Answers

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Light of 500nm is incident on a metal with work function 2.28ev .The de broglie wavelength of the emitted electron is 1) less than 2.8 ×10-12m 2)less than 2.8×10-10m 3) less than 2.8×10-9m 4) greater than 2.8 ×109m

Light of 500nm is incident on a metal with work function 2.28ev .The de broglie wavelength of the emitted electron is
1) less than 2.8 ×10-12m

2)less than 2.8×10-10m
3) less than 2.8×10-9m
4) greater than 2.8 ×109m