SATHEE: Modern Physics 7 Question 51

Modern Physics 7 Question 51

57. A gas of identical hydrogen-like atoms has some atoms in the lowest (ground) energy level $A$ and some atoms in a particular upper (excited) energy level $B$ and there are no atoms in any other energy level. The atoms of the gas make the transition to a higher energy level by absorbing monochromatic light of photon energy $2.7 e V$ . Subsequently, the atoms emit radiation of only six different energy photons. Some of the emitted photons have an energy of $2.7 e V$ , some have more energy and some less than $2.7 e V$ .

$(1989, 8 M)$

(a) Find the principal quantum number of the initially excited level $B$ .

(b) Find the ionization energy for the gas atoms.

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Answer:

Correct Answer: 57. (a) 2 (b) $14.4 e V$ (c) $13.5 e V, 0.7 e V$

Solution:

(a) In emission spectrum total six lines are obtained. Hence, after excitation if $n_{f}$ be the final principal quantum number then,

$\frac{n_{f} (n_{f} - 1)}{2} = 6 or n_{f} = 4 z$

i.e. after excitation atom goes to $4^{th}$ energy state. Hence, $n_{i}$ can be either 1,2 or 3 .

Absorption and emission spectrum corresponding to $n_{i} = 1, n_{i} = 2$ and $n_{i} = 3$ are shown in figure.

For $n_{i} = 1$ , energy of emitted photons $\leq 2.7 e V$

For $n_{i} = 2$ , energy of emitted photons $>=< 2.7 e V$ and

For $n_{i} = 3$ , energy of emitted photons $\geq 2.7 e V$ .

As per the given condition $n_{i} = 2$ .

(b)

$E_{4} - E_{2} = 2.7 e V$

$\frac{E_{1}}{16} - \frac{E_{1}}{4} = 2.7$

$\begin{aligned} - \frac{3}{16} E_{1} & = 2.7 \\ E_{1} & = - 14.4 e V \end{aligned}$

Therefore, ionization energy for the gas atoms is $| E_{1} |$ or $14.4 e V$ .

$\begin{aligned} ∴ E_{max} & = E_{4} - E_{1} \\ = \frac{E_{1}}{16} - E_{1} = \frac{- 15 E_{1}}{16} \\ = - \frac{15}{16} (14.4) = 13.5 e V \end{aligned}$

Similarly, minimum energy of the emitted photons is corresponding to transition from $n = 4$ to $n = 3$ .

$\begin{aligned} ∴ E_{min} & = E_{4} - E_{3} \\ = \frac{E_{1}}{16} - \frac{E_{1}}{9} = - \frac{7 E_{1}}{16 \times 9} \\ = \frac{7 \times 14.4}{16 \times 9} = 0.7 e V \end{aligned}$