Question

A hydrogen atom in ground state absorbs a photon of ultraviolet radiation of wavelength 50 nm. Assuming that the entire photon energy is taken up by the electron with what kinetic energy will the electron be ejected?

Answer 1

The kinetic energy of the ejected electron will be K.E = 11.24 eV

Explanation:

λ = 50 nm

Work function = Energy required to remove the electron from n1 = 1 to n2 = ∞

E = 13.6 (1/1 - 1/∞)  = 13.6

hc / λ  - 13.6 = K.E

1242 /50 - 13.6 = K.E

K.E = 24.84 - 13.6

K.E = 11.24 eV

Thus the kinetic energy of the ejected electron will be K.E = 11.24 eV

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Answer 2

The electron will be ejected with 11.24 eV of kinetic energy when the entire photon energy is taken in by the electron.

EXPLANATION:

It is given that:

Ultraviolet radiation’s wavelength, λ = 50 nm.  

It is understood that the atom’s work function is the energy that is necessary for removing an electron from the atom’s surface. Hence, the work function can be found by finding the energy necessary to eradicate the electron to

$n_2 = \infty from n_1 = 1.$

Work function,$W_0 = 13.6(\frac{1}{1} - \frac{1}{\infty} ) = 13.6 eV$

We will obtain photoelectric equation of Einstein, we get  

$E = W_0 + KE$

$KE = E - W_0$

KE = 24.84 − 13.6

KE = 11.24 eV.

Therefore the Kinetic energy ejected by the electron is 11.24 eV.