Ahydrogen 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.

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;

ΔE = P*C

n₁ = 7

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

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures