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Physics Question on Atoms

Given the value of Rydberg constant is 107 m−110^7 \, m^{-1}, the wave number of the last line of the Balmer series in hydrogen spectrum will be :

A

0.5×107 m−10.5 \times 10^7 \, m^{-1}

B

0.25×107 m−10.25 \times 10^7 \, m^{-1}

C

2.5×107 m−12.5 \times 10^7 \, m^{-1}

D

0.025×104 m−10.025 \times 10^4 \, m^{-1}

Answer

0.25×107 m−10.25 \times 10^7 \, m^{-1}

Explanation

Solution

1λ=RZ2(1n22−1n12)=107×12(122−1∞2)\frac{1}{\lambda } = RZ^{2} \left(\frac{1}{n_{2}^{2}} - \frac{1}{n_{1}^{2}}\right) = 10^{7} \times1^{2} \left(\frac{1}{2^{2}} - \frac{1}{ \infty^{2}}\right)
⇒\Rightarrow wave number =1λ=0.25×107m−1= \frac{1}{\lambda } = 0.25 \times10^{7} m^{-1}