Question

When a hydrogen atom is excited from ground state to first excited state then

• A. its kinetic energy increases by $10.2eV$
• B. its kinetic energy decreases by $10.2eV$
• C. its potential energy increases by $20.4eV$
• D. its angular momentum increases by $1.05\times {10}^{-34}Js$

Answer 1

Answer: (B), (C), (D)

The correct options are
B its kinetic energy decreases by $10.2eV$
C its potential energy increases by $20.4eV$
D its angular momentum increases by $1.05\times {10}^{-34}Js$

For ground state $n=1$
First excited state $n=2$

By centripetal force balance on electron:
$\frac{k{e}^2}{r^2}=\frac{m{v}^2}{r}$
$\Rightarrow 2KE=-U$

Total energy, $E=KE+U=-KE$
We know, $E=-13.6\frac{1}{n^2}eV$
Transition from ground state to first exicted state:
$(n=1)\to (n=2)$
$\Delta E=-13.6(\frac{1}{2^2}-\frac{1}{1^2})=10.2eV$

$\Delta KE=-10.2eV$
$\Delta U=-2KE=20.4eV$

Angular momentum,
$L=mvr=\frac{nh}{2\pi }$
$\Delta L=\frac{\Delta nh}{2\pi }=\frac{6.6\times {10}^{-34}}{6.28}$
$\Delta L=1.05\times {10}^{-34}Js$