Question

A hydrogen atom emits a photon that has momentum 0.3059×10^(-27) kg·m/s. This photon is emitted because the electron in the atom falls from a higher energy level into the n = 4 level. What is the quantum number of the level from which the electron falls? Use values of h = 6.626×10^-34 J·s, c = 2.998×10^8 m/s, and e = 1.602×10^(-19) C.

Answer 1

The quantum number of the higher energy level is 7

Step-by-step explanation:

Given:

Momentum (p) = 0.3059×10⁻²⁷ kg·m/s

Planck's constant (h) = 6.626×10⁻³⁴ J·s

Speed of light (c) = 2.998×10⁸m/s

Charge of electron (e) = 1.602×10⁻¹⁹ C

Lower energy level (n₂) = 4

Higher energy level (n₁) = ?

From Bohr's model, change in energy of a photon is given as;

`\delta E= ((1)/((n_2)^2) -(1)/((n_1)^2))*13.6eV`

ΔE = P*C

`P*C = ((1)/((n_2)^2) -(1)/((n_1)^2))*13.6eV`

`(P*C)/(13.6eV) =(1)/((n_2)^2) -(1)/((n_1)^2)`

`(1)/((n_1)^2)=(1)/((n_2)^2) -(P*C)/(13.6eV)`

`(1)/((n_1)^2)=(1)/((4)^2) -(0.3059X10^(-27)*2.998X10^8)/(13.6X1.602X10^(-19))`

`(1)/((n_1)^2)=(1)/((16)) -(0.9177)/(21.7812)}`

`(1)/((n_1)^2)=0.0625 -0.0421`

`(1)/((n_1)^2)=0.0204`

`(n_1)^2 = (1)/(0.0204)`

`(n_1)^2 = 49.0196`

`n_1 =√(49.0196)`

n₁ = 7

Therefore, the quantum number of the higher energy level is 7