Question

In hydrogen atom, electron is present in $n_x$ state. The energy of Lyman spectral line of hydrogen spectrum originated from $n_x$ state is $1.635 \times 10^{-18}\text{ J}$. What is the approximate energy (in J) required to excite this electron from $n_x$ state to $(n_x+1)$ state?}

• A. $3 \times 10^{-19}$
• B. $3 \times 10^{-18}$
• C. $1.6 \times 10^{-18}$
• D. $3 \times 10^{-20}$

Hint:

The value $1.635\times10^{-18}\text{ J}$ is exactly $\frac{3}{4}$ of $2.18\times10^{-18}\text{ J}$, immediately giving $\frac{1}{n^2}=\frac14$ and hence $n=2$.

Answer 1

The Correct Option is A

Concept: The energy of an electron in the $n^{\text{th}}$ orbit of a hydrogen atom is \[ E_n=-\frac{2.18\times10^{-18}}{n^2}\text{ J} \] For a Lyman series transition from $n_x$ to $n=1$, the emitted energy is \[ \Delta E=2.18\times10^{-18}\left(1-\frac{1}{n_x^2}\right)\text{ J} \]

Step 1: Determine the value of $n_x$ Given, \[ 1.635\times10^{-18} = 2.18\times10^{-18} \left(1-\frac{1}{n_x^2}\right) \] Dividing both sides by $2.18\times10^{-18}$, \[ \frac{1.635}{2.18} = 1-\frac{1}{n_x^2} \] \[ 0.75 = 1-\frac{1}{n_x^2} \] \[ \frac{1}{n_x^2} = 0.25 = \frac{1}{4} \] \[ n_x=2 \]

Step 2: Calculate the excitation energy from $n=2$ to $n=3$ \[ \Delta E = 2.18\times10^{-18} \left( \frac{1}{2^2}-\frac{1}{3^2} \right) \] \[ = 2.18\times10^{-18} \left( \frac{1}{4}-\frac{1}{9} \right) \] \[ = 2.18\times10^{-18} \times\frac{5}{36} \] \[ \Delta E \approx 3.03\times10^{-19}\text{ J} \] Therefore, \[ \boxed{\Delta E\approx3\times10^{-19}\text{ J}} \]