Question

Given the value of Rydberg constant is ${10}^7{m}^{-1}$, the waves number of the last line of the Balmer series in hydrogen spectrum will be:

• A. $0.025\times {10}^4{m}^{-1}$
• B. $0.5\times {10}^7{m}^{-1}$
• C. $0.25\times {10}^7{m}^{-1}$
• D. $2.5\times {10}^7{m}^{-1}$

Answer 1

The correct option is C

$0.25\times {10}^7{m}^{-1}$

Step 1: Given

Rydberg constant: $R={10}^7{m}^{-1}$

Step 2: Formula Used

Rydberg formula for wave number is given as $\overline{\nu }=\frac{1}{\lambda }=R\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)$, where $\overline{\nu }$ is wave number, $\lambda$ is wavelength, $n_1=2$ is the number of first-line which is equal to $2$ for Balmer series and $n_2=\infty$ for last spectral line of Balmer series.

Step 3: Solution

Substitute the values to find the wave number.

$\begin{array}{rcl}\overline{\nu }& =& R\left(\frac{1}{{n_1}^2}-\frac{1}{{n_2}^2}\right)\\ & =& {10}^7\left(\frac{1}{2^2}-\frac{1}{{\infty }^2}\right)\left[\frac{1}{\infty }=0\right]\\ & =& {10}^7\left(\frac{1}{4}-0\right)\\ & =& {10}^7\times \frac{1}{4}\\ & =& 0.25\times {10}^{7}m\end{array}$

Hence, the correct option is option C.