Question

An electron in a hydrogen atom drops from the n=3 energy level to the n=2 energy level

Answer 1

Answer:

We have the change in energy levels of the electron to be - 0.302 x 10⁻¹⁸ J.

Explanation:

Let us consider the first orbit to be n=3, and the second orbit to be n=2.

For the energy level to drop, the electron's potential energy must change into its kinetic energy, according to the law of conservation of energy.

Post derivation, we get the formula to be -

$E = R_h (\frac{1}{n_1^2} - \frac{1}{n_2^2} )$

Now, we have n₁ to be 3, and n₂ to be 2.

Also, Rh is Rydberg's constant, i.e., 2.18 x 10⁻¹⁸.

Thus, substituting the values in the formula, we get

$E = 2.18 \times 10^-^1^8 (\frac{1}{(3)^2} -\frac{1}{(2)^2} )\\\\E = 2.18 \times 10^-^1^8 (\frac{1}{9} -\frac{1}{4} )\\\\E = 2.18 \times 10^-^1^8 (\frac{4-9}{36} )\\E = \frac{-10.9 \times 10^-^1^8}{36}\\E = -0.302 \times 10^-^1^8 J$

Thus, we have the change in energy levels to be - 0.302 x 10⁻¹⁸ J.

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