Modern Physics 1 Question 51
47. 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)
Show Answer
Answer:
Correct Answer: 47. $n=2, Z=4,-217.6 eV, 10.58 eV$
Solution:
- Let ground state energy (in $eV$ ) be $E _1$.
Then, from the given condition
$$ \begin{aligned} & E _{2 n}-E _1=204 eV \\ & \text { or } \\ & \frac{E _1}{4 n^{2}}-E _1=204 eV \\ & \text { or } \quad E _1 \frac{1}{4 n^{2}}-1=204 eV \\ & \text { and } \quad E _{2 n}-E _n=40.8 eV \\ & \text { or } \quad \frac{E _1}{4 n^{2}}-\frac{E _1}{n^{2}}=40.8 eV \\ & \text { or } \quad E _1 \frac{-3}{4 n^{2}}=40.8 eV \end{aligned} $$
From Eqs. (i) and (ii),
$$ \begin{gathered} \frac{1-\frac{1}{4 n^{2}}}{\frac{3}{4 n^{2}}}=5 \\ \text { or } \quad 1=\frac{1}{4 n^{2}}+\frac{15}{4 n^{2}} \text { or } \frac{4}{n^{2}}=1 \\ \text { or } \quad n=2 \end{gathered} $$
From Eq. (ii),
$$ \begin{aligned} E _1 & =-\frac{4}{3} n^{2}(40.8) eV \\ & =-\frac{4}{3}(2)^{2}(40.8) eV \\ E _1 & =-217.6 eV \\ E _1 & =-(13.6) Z^{2} \\ Z^{2} & =\frac{E _1}{-13.6}=\frac{-217.6}{-13.6}=16 \\ Z \quad \quad \quad E _{\min } & =E _{2 n}-E _{2 n-1} \\ & =\frac{E _1}{4 n^{2}}-\frac{E _1}{(2 n-1)^{2}} \\ & =\frac{E _1}{16}-\frac{E _1}{9}=-\frac{7}{144} E _1 \\ & =-\frac{7}{144}(-217.6) eV \\ \therefore \quad E _{\min } & =10.58 eV \end{aligned} $$
