Modern Physics Ques 60
- A hydrogen like atom of atomic number $Z$ is in an excited state of quantum number $2 $ $n$. It can emit a maximum energy photon of $204$ $ eV$. If it makes a transition to quantum state $n$, a photon of energy $40.8$ $ eV$ is emitted. Find $n, Z$ and the ground state energy (in $eV$ ) of this atom. Also, calculate the minimum energy (in $eV$ ) that can be emitted by this atom during de-excitation. Ground state energy of hydrogen atom is $-13.6$ $ eV$.
$(2000,6$ M)
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Answer:
Correct Answer: 60.$n=2, Z=4,-217.6 $ $eV, 10.58$ $ eV$
Solution:
Formula:
- Let ground state energy (in $eV$ ) be $E _1$.
Then, from the given condition
$ \begin{aligned} & E _{2 n}-E _1=204 eV \\ & \text { or } \\ & \frac{E _1}{4 n^{2}}-E _1=204 eV \\ & \text { or } \quad E _1 (\frac{1}{4 n^{2}}-1)=204 eV \quad …….(i)\\ & \text { and } \quad E _{2 n}-E _n=40.8 eV \\ & \text { or } \quad \frac{E _1}{4 n^{2}}-\frac{E _1}{n^{2}}=40.8 eV \\ & \text { or } \quad E _1 (\frac{-3}{4 n^{2}})=40.8 eV \quad …….(ii) \end{aligned} $
From Eqs. (i) and (ii),
$\big(\frac{1-\frac{1}{4 n^{2}}}{\frac{3}{4 n^{2}}})=5 $
$\text { or } \quad 1=\frac{1}{4 n^{2}}+\frac{15}{4 n^{2}} \text { or } \frac{4}{n^{2}}=1 $
$\text { or } \quad n=2$
From Eq. (ii),
$ \begin{aligned} E _1 & =-\frac{4}{3} n^{2}(40.8) eV \\ & =-\frac{4}{3}(2)^{2}(40.8) eV \\ E _1 & =-217.6 eV \\ E _1 & =-(13.6) Z^{2} \\ Z^{2} & =\frac{E _1}{-13.6}=\frac{-217.6}{-13.6}=16 \\ Z \quad \quad \quad E _{\min } & =E _{2 n}-E _{2 n-1} \\ & =\frac{E _1}{4 n^{2}}-\frac{E _1}{(2 n-1)^{2}} \\ & =\frac{E _1}{16}-\frac{E _1}{9}=-\frac{7}{144} E _1 \\ & =-(\frac{7}{144})(-217.6) eV \\ \therefore \quad E _{\min } & =10.58 eV \end{aligned} $
BETA
