Question

Hydrogen atoms are excited from n=1 →n=2. The electron in the n=2 shell then returns to n=1 emitting as photon. These emitted photons hit a piece of copper (work function = 2.40 eV) causing electron s to be ejected. Calculate the energy of these ejected electrons in joules. (A) 1.63 x 10-18 J (B) 0.384 x 10-18 j (C) 2.18 x 10-18 j (D) 1.25 x 10-18 J​

Answer 1

option 2 is the correct answer

Answer 2

Answer:

The energy of ejected electrons from copper surface is equal to 1.25 ×10⁻¹⁸J.

Therefore, option (D) is correct.

Explanation:

We know that the energy of Hydrogen atom for nth level can be calculated from the formula:

$E _n =-\frac{13.6}{n^2}\;eV$

So the energy of electron in n = 1 , $E_1 = -13.6eV$

The energy of n = 2, $E_2 = \frac{-13.6}{(2)^2} = -3.4eV$

The energy difference in the n = 2 and n = 1 energy level:-

$E_2-E_1 =-3.4-(-13.6) = -3.4 + 13.6 = 10.2eV$

Therefore, the emitted emitted has energy equals to 10.2eV.

Given, the work function = 2.4eV when this emitted photon hit a piece of copper.

Work function is the minimum amount of energy is required to eject the electron from the metal surface.

Energy of ejected electrons = Energy of photon - work function

Energy of ejected electrons from copper $= 10.2eV - 2.4 eV = 7.8eV$

To convert the electron volt in Joules, $1eV = 1.6\times 10^{-19}J$

Energy of electrons $= 7.8 \times 1.6\times 10^{-19}J =1.25 \times 10^{-18}J$

Therefore, the energy of these ejected electrons is equal to 1.25 ×10⁻¹⁸J.

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