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Question:
The minimum energy needed to remove an electron from a metal corresponds to a wavelength of 500 nm. What is the total kinetic energy of all the photoelectrons ejected per second when the entire radiation from a 100 Watt bulb with a wavelength of 300 nm falls on the surface of the metal?
[Planck's constant = 6.6 x 10^-34 J s; speed of light = 3 x 10^8 m s-1]
(a) 40J
(b) 2.6x10-19
(c) 1.6x10-19
(d) 80J
​

Answer :

yashikayadav5775

Explanation:

To solve this problem, we need to calculate the total kinetic energy of all the photoelectrons ejected per second when the radiation from the 100 Watt bulb falls on the metal surface.

Given:

- Minimum energy to remove an electron from the metal corresponds to a wavelength of 500 nm.

- Wavelength of radiation from the 100 Watt bulb = 300 nm.

- Planck's constant (h) = 6.6 x 10^-34 J s.

- Speed of light (c) = 3 x 10^8 m/s.

First, calculate the energy (E) of a photon with a wavelength of 500 nm using the formula:

E = hc / λ

where:

h = Planck's constant

c = speed of light

λ = wavelength in meters

Next, calculate the energy of a photon with a wavelength of 300 nm using the same formula.

Then, find the difference in energy between the two photons, as this energy will be used to eject electrons from the metal.

Finally, since each photon can eject one electron, the total kinetic energy of all the photoelectrons ejected per second is equal to the energy difference calculated above multiplied by the power of the bulb (100 Watts).

Perform these calculations step by step, and you should be able to determine the correct answer. If you need further assistance with the calculations, feel free to ask!

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