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

The Correct Option is C

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 E_initial = 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 = E_initial - E_final = (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.