Modern Physics Ques 65
- The ionization energy of a hydrogen-like Bohr atom is $13.6$ eV.
(1984, 4M)
(a) What is the wavelength of the radiation emitted when the electron jumps from the first excited state to the ground state?
(b) What is the radius of the first orbit for this atom?
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Answer:
Correct Answer: 65.(a) $300$ $ \AA \quad $ (b) $0.2645 $ $ \AA$
Solution:
Formula:
- (a) 1 rydberg $=2.2 \times 10^{-18} J=R h c$
Ionisation energy is given as $4$ eV
$ =8.8 \times 10^{-18} J=\frac{8.8 \times 10^{-18}}{1.6 \times 10^{-19}}=55 $ $eV $
$\therefore \quad$ Energy in first orbit $E _1=-13.6 $ $eV$
Energy of radiation emitted when electron jumps from first excited state $(n=2)$ to ground state $(n=1)$ :
$ E _{21}=\frac{E _1}{(2)^{2}}-E _1=-\frac{3}{4} E _1=41.25\ eV $
$\therefore$ Wavelength of photon emitted in this transition is
$ \lambda=\frac{12375}{41.25}=300 $ $\AA $
(b) Let $Z$ be the atomic number of given element. Then
$E _1 =(-13.6)\left(Z^{2}\right) $
$\text { or } \quad -55 =(-13.6)\left(Z^{2}\right) $
$\text { or } \quad Z \approx 2 $
$ \text { Now, as } r \propto \frac{1}{Z} \text{ for hydrogen-like atoms}$
Radius of first orbit of this atom,
$ r _1=\frac{r _{H _1}}{Z}=\frac{0.529}{2}=0.2645 $ $ \AA $
BETA
