The figure shows the energy level diagram of the hydrogen atom. Out of the six given transitions, which transition results in the emission of a photon of wavelength 4870 Å?: ($h = 6.63 × 10^{-34} Js$)

n = 4 to n = 2

The correct answer is Option (1) → n = 4 to n = 2

$\text{Given wavelength: } \lambda = 4870 \ \text{Å} = 487 \times 10^{-9}\,\text{m}.$

$E = \frac{hc}{\lambda} = \frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{487 \times 10^{-9}}$

$= 4.09 \times 10^{-19}\,\text{J} = \frac{4.09 \times 10^{-19}}{1.6 \times 10^{-19}} \,\text{eV}$

$= 2.55 \,\text{eV}$

$\text{Now, energy difference for transition } n=4 \to n=2:$

$E = (-3.4) - (-0.85) = 2.55 \,\text{eV}$

$\Rightarrow \ \text{This exactly matches the photon energy.}$

${\ \text{Correct transition is } n=4 \to n=2\ }$