Answer:
The quantum number of the higher energy level is 7
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;
[tex]\delta E= (\frac{1}{(n_2)^2} -\frac{1}{(n_1)^2})*13.6eV[/tex]
ΔE = P*C
[tex]P*C = (\frac{1}{(n_2)^2} -\frac{1}{(n_1)^2})*13.6eV[/tex]
[tex]\frac{P*C}{13.6eV} =\frac{1}{(n_2)^2} -\frac{1}{(n_1)^2}[/tex]
[tex]\frac{1}{(n_1)^2}=\frac{1}{(n_2)^2} -\frac{P*C}{13.6eV}[/tex]
[tex]\frac{1}{(n_1)^2}=\frac{1}{(4)^2} -\frac{0.3059X10^{-27}*2.998X10^8}{13.6X1.602X10^{-19}}[/tex]
[tex]\frac{1}{(n_1)^2}=\frac{1}{(16)} -\frac{0.9177}{21.7812}}[/tex]
[tex]\frac{1}{(n_1)^2}=0.0625 -0.0421[/tex]
[tex]\frac{1}{(n_1)^2}=0.0204[/tex]
[tex](n_1)^2 = \frac{1}{0.0204}[/tex]
[tex](n_1)^2 = 49.0196[/tex]
[tex]n_1 =\sqrt{49.0196}[/tex]
n₁ = 7
Therefore, the quantum number of the higher energy level is 7
