Question

When an electron in hydrogen atom is excited, from its $4^{th}$ to $5^{th}$ stationary orbit, the change in angular momentum of electron is (Planck’s constant:$h=6.6\times {10}^{-34}J-s$)

• A. $4.16\times {10}^{-34}J-s$
• B. $3.32\times {10}^{-34}J-s$
• C. $1.05\times {10}^{-34}J-s$
• D. $2.08\times {10}^{-34}J-s$

Answer 1

The correct option is C $1.05\times {10}^{-34}J-s$

Change in angular momentum
$\Delta L=L_2-L_1=\frac{n_{2}h}{2\pi }-\frac{n_{1}h}{2\pi }\Rightarrow \Delta L=\frac{h}{2\pi }(n_2-n_1)=\frac{6.6\times {10}^{-34}}{2\times 3.14}(5-4)=1.05\times {10}^{-34}J-s$