Question

The wave number of first line in Balmer series of hydrogen is $15200{cm}^{-1}$. What is the wave number of first line in Balmer series of ${Be}^{3+}$?

• A. $2.43\times {10}^5{cm}^{-1}$.
• B. $2.43\times {10}^6{cm}^{-1}$.
• C. $8\times {10}^5{cm}^{-1}$.
• D. $8\times {10}^6{cm}^{-1}$.

Answer 1

The correct option is A $2.43\times {10}^5{cm}^{-1}$.

$Explanation:$

According to $RydbergEquation$

$\overrightarrow{\nu }=\frac{1}{\lambda }=R_H\times Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

Where,

$\overrightarrow{\nu }=$ Wave number

$\lambda =$ Wave length

$R_H=$ Rydberg constant

$Z=$ Atomic number

$n_1=$ Initial energy level

$n_2=$ Final energy level

For 1st line of $Balmerseries$, $n_1=2,n_2=3$

For Hydrogen atom:

${\overrightarrow{\nu }}_H=\frac{1}{{\lambda }_H}=R_H\times Z^2[\frac{1}{2_2}-\frac{1}{3_2}]=15200c{m}^{-1}$

For Beryllium ion $\left(B{e}^{+3}\right):$

${\overrightarrow{\nu }}_{Be}=\frac{1}{{\lambda }_{Be}}=R_H\times 4^2[\frac{1}{2_2}-\frac{1}{3_2}]$

${\overrightarrow{\nu }}_{Be}=16\times {\overrightarrow{\nu }}_H=16\times 15200=243200c{m}^{-1}=2.43\times {10}^{5}c{m}^{-1}$

${\overrightarrow{\nu }}_{Be}=2.43\times {10}^{5}c{m}^{-1}$

Hence the correct answer is option $\left(A\right)$.