Modern Physics 1 Question 58
54. Hydrogen atom in its ground state is excited by means of monochromatic radiation of wavelength $975^{\circ} \AA$. How many different lines are possible in the resulting spectrum? Calculate the longest wavelength amongst them. You may assume the ionization energy for hydrogen atom as $13.6 eV$.
$(1982,5 M)$
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Answer:
Correct Answer: 54. Six, $1.875 \mu m$
Solution:
- Energy corresponding to given wavelength
$$ E(\text { in } eV)=\frac{12375}{\lambda(\text { in } \AA)}=\frac{12375}{975}=12.69 eV $$
Now, let the electron excites to $n^{\text {th }}$ energy state. Then,
$$ \begin{aligned} & E _n-E _1=12.69 \text { or } \frac{(-13.6)}{\left(n^{2}\right)}-(-13.6)=12.69 \\ & \therefore \quad n \approx 4 \end{aligned} $$
i.e. electron excites to 4 th energy state. Total number of lines in emission spectrum would be
$$ \frac{n(n-1)}{2}=\frac{4 \times 3}{2}=6 $$
Longest wavelength will correspond to the minimum energy and minimum energy is released in transition from $n=4$ to $n=3$.
$$ E _{4-3}=E _4-E _3=\frac{-13.6}{\left(4^{2}\right)}-\frac{-13.6}{(3)^{2}}=0.66 eV $$
$\therefore$ Longest wavelength will be,
$$ \begin{aligned} \lambda _{\max } & =\frac{12375}{E(\text { in } eV)} \\ & =\frac{12375}{0.66} \AA=1.875 \times 10^{-6} m=1.875 \mu m \end{aligned} $$
