Question

find the no. of electrons emitted per second by a 24W source of monochromatic light of wavelength 6600 angstrom, assuming 3 percent efficiency for photoelectric effect.

Answer 1

When the cathode is exposed to light, electrons are ejected from its photosensitive surface. These electrons are attracted to the positive anode and form a current that can be measured with an electrometer. where h = Planck's constant, f = frequency of the light, and Φ is the work function of the cathode surface.

Answer 2

The number of photons emitted per second are $n=2.4\times 10^{18}\ photons$

Explanation:

It is given that,

Power of the source, P = 24 W

Wavelength, $\lambda=6600\times 10^{-10}\ m$

Energy of a photon is given by :

$E=\dfrac{hc}{\lambda}$

$E=\dfrac{6.6\times 10^{-34}\times 3\times 10^8}{6600\times 10^{-10}}$

$E=3\times 10^{-19}\ J$

If n is the number of photons incoming is given by :

$n=\dfrac{24}{3\times 10^{-19}}$

$n=8\times 10^{19}\ photons$

If efficiency is of 3 percent. So, number of photons are :

$n=\dfrac{3}{100}\times 8\times 10^{19}$

$n=2.4\times 10^{18}\ photons$

So, the number of photons emitted per second is $2.4\times 10^{18}\ photons$. Hence, this is the required solution.

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Photoelectric effect

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