Question

The atomic number (Z) of a hydrogen like atom whose shortest wavelength of Brackett Series is same as the shortest wavelength of Balmer series of hydrogen atom is

• A. Z = 1
• B. Z = 2
• C. Z = 3
• D. Z = 4

Hint:

Shortest wavelength always means $n_{2} = \infty$. The equation simplifies directly to balancing $\frac{Z^2}{n_{1}^2}$ values between the two configurations.

Answer 1

The Correct Option is B

Step 1: Rydberg formula.
For a hydrogen-like atom $\dfrac{1}{\lambda} = RZ^{2}\left(\dfrac{1}{n_{1}^{2}} - \dfrac{1}{n_{2}^{2}}\right)$.

Step 2: Shortest wavelength meaning.
The shortest wavelength (series limit) is when $n_{2} = \infty$. Then the second term is zero.

Step 3: Balmer limit of hydrogen.
Balmer has $n_{1}=2$ and $Z=1$: \[ \frac{1}{\lambda_{B}} = R(1)\frac{1}{2^{2}} = \frac{R}{4} \]
Step 4: Brackett limit of the unknown atom.
Brackett has $n_{1}=4$: \[ \frac{1}{\lambda_{Br}} = RZ^{2}\frac{1}{4^{2}} = \frac{RZ^{2}}{16} \]
Step 5: Set the two equal.
\[ \frac{R}{4} = \frac{RZ^{2}}{16} \quad\Rightarrow\quad 4 = Z^{2} \]
Step 6: Solve for $Z$.
\[ Z = 2 \]So the atomic number is $2$, which is option 2.
\[ \boxed{Z = 2} \]