Question

In a transition to a state of excitation energy $10.19\text{eV},$ hydrogen atom emits a photon of wavelength $4890\mathring{\text{A}}.$ The Binding energy of the initial state is nearly equal to

• A. $1.5\text{eV}$
• B. $3.4\text{eV}$
• C. $0.5\text{eV}$
• D. $0.9\text{eV}$

Answer 1

The correct option is D $0.9\text{eV}$

Given,

$\lambda =4890\mathring{\text{A}}\text{Excitation energy}=10.19\text{eV}$

Energy of emitted photon is,

$E=\frac{hc}{\lambda }=\frac{12400}{4890}=2.54\text{eV}$

The excitation energy is the energy to excite the atom to a level above the ground state. Therefore, the energy level is,

$E=–13.6+10.19=–3.41\text{eV}$

Photon arises from transition between energy state such that,

$E_i-E_f=hf=2.54\text{eV}$

$E_i=2.54+E_f$

$E_i=2.54+\left(E\right)=2.54-3.41=-0.87\text{eV}$

$\therefore \text{Binding energy of the initial state}\approx 0.9\text{eV}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(D\right)$ is the correct answer.