Question

The energy corresponding to one of the lines in the Paschen series for $He^+$ ion is $4.25 \times 10^{-19}$ J. Find the principle quantum number for the transition which produces this line. (Take $R_H = 1.1 \times 10^7 m^{-1}$)

Answer 1

4

Answer 2

For a hydrogen-like ion, the Rydberg formula is

$$\frac{1}{\lambda}=Z^2R_H\left(\frac{1}{n_f^2}-\frac{1}{n_i^2}\right)$$

In the Paschen series, the final level is $n_f=3$ and for $He^+$ the nuclear charge is $Z=2$.

The energy of the photon is given by

$$E=\frac{hc}{\lambda}\quad \Longrightarrow\quad \frac{E}{hc} = \frac{1}{\lambda}$$

Substitute into the Rydberg formula:

$$\frac{E}{hc}=Z^2R_H\left(\frac{1}{3^2}-\frac{1}{n_i^2}\right)$$

Plug in the values:

$$\frac{4.25\times10^{-19}}{6.626\times10^{-34}\times3\times10^8} = 4\times(1.1\times10^7)\left(\frac{1}{9}-\frac{1}{n_i^2}\right)$$

Compute the left side (noting that $6.626\times 3$ approximates a factor used in the hint):

$$\frac{4.25\times10^{-19}}{1.9878\times10^{-25}} \approx 2.14\times10^6 \, \text{m}^{-1}$$

Now, the right side:

$$4\times1.1\times10^7 = 4.4\times10^7 \, \text{m}^{-1}$$

Thus,

$$2.14\times10^6 = 4.4\times10^7\left(\frac{1}{9}-\frac{1}{n_i^2}\right)$$

Solving for the term in parentheses:

$$\frac{1}{9}-\frac{1}{n_i^2} = \frac{2.14\times10^6}{4.4\times10^7} \approx 0.0486$$

So,

$$\frac{1}{n_i^2} = \frac{1}{9} - 0.0486 \approx 0.1111 - 0.0486 = 0.0625$$

Finally,

$$n_i^2 = \frac{1}{0.0625} = 16 \quad \Longrightarrow \quad n_i=4.$$

Core Explanation

• Use the Rydberg formula for hydrogen-like ions.
• For Paschen series, $n_f=3$, and for $He^+$, $Z=2$.
• Equate the expression for $\frac{E}{hc}$ with the Rydberg formula and solve for $n_i$.
• Calculation yields $n_i = 4$.

The transition is from $n_i=4$ to $n_f=3$.