Question

Energy E of a hydrogen atom with principal quantum number n is given by E = $-\frac{13.6}{n^2}eV$. The energy of a photon ejected when the electron jumps from the n = 3 state to the n = 2 states of hydrogen is approximately

• A. 0. 85 eV
• B. 3.4 eV
• C. 1.9 eV
• D. 1.5 eV

Answer 1

Answer: (C)

1.9 eV

Answer 2

The energy of the photon emitted when an electron jumps from a higher energy level to a lower energy level in a hydrogen atom can be calculated using the formula: ΔE = E_initial - E_final In this case,
the initial energy level (n = 3) corresponds to Einitial = 2$\times$ 32$\times$(-13.6 eV),
and the final energy level (n = 2) corresponds to E_final = 2$\times$22$\times$(-13.6 eV).
Now, we can calculate the energy difference: ΔE = Einitial - Efinal = (2$\times$32$\times$(-13.6 eV)) - (2$\times$22$\times$(-13.6 eV)) ΔE
= (2$\times$9$\times$(-13.6 eV)) - (2$\times$4$\times$(-13.6 eV)) ΔE
= (-244.8 eV) - (-108.8 eV) ΔE
= -136 eV
So, the energy of the photon ejected when the electron jumps from n = 3 to n = 2 states of hydrogen is approximately 136 eV.