Question

In H atom, electron is present in $n_x$ state. The angular momentum of this electron is $1.051 \times 10^{-34}\text{ Js}$. What is the energy (in J) required to excite this electron from $n_x$ state to $(n_x+1)$ state? ($h = 6.6 \times 10^{-34}\text{ Js}$; $\pi = 3.14$)

• A. $2.18 \times 10^{-18}$
• B. $0.545 \times 10^{-18}$
• C. $1.93 \times 10^{-19}$
• D. $1.635 \times 10^{-18}$

Hint:

For the first Bohr orbit, the angular momentum is $\frac{h}{2\pi}\approx1.05\times10^{-34}\text{ Js}$. Recognizing this value instantly identifies $n=1$.

Answer 1

The Correct Option is D

Concept: According to Bohr's quantization condition, \[ L=\frac{nh}{2\pi} \] where $L$ is the angular momentum.

Step 1: Find the quantum number $n_x$ Given, \[ 1.051\times10^{-34} = \frac{n_x(6.6\times10^{-34})}{2(3.14)} \] \[ 1.051 = \frac{6.6}{6.28}n_x \] \[ 1.051 = 1.051\,n_x \] \[ n_x=1 \] Thus the electron is in the ground state.

Step 2: Calculate energy required for excitation from $n=1$ to $n=2$ \[ \Delta E = 2.18\times10^{-18} \left( \frac{1}{1^2}-\frac{1}{2^2} \right) \] \[ = 2.18\times10^{-18} \left( 1-\frac14 \right) \] \[ = 2.18\times10^{-18} \times\frac34 \] \[ = 1.635\times10^{-18}\text{ J} \] Hence, \[ \boxed{1.635\times10^{-18}\text{ J}} \]