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Physics Question on Electromagnetic Spectrum

The limiting line in Balmer series will have a frequency of (Rydberg constant, R∞=3.29×1015R_{\infty}=3.29\times10^{15} cycles/s)

A

8.22×1014s−18.22 \times 10^{14} s^{-1}

B

3.29×1015s−13.29 \times 10^{15} s^{-1}

C

3.65×1014s−13.65 \times 10^{14} s^{-1}

D

5.26×1013s−15.26 \times 10^{13} s^{-1}

Answer

8.22×1014s−18.22 \times 10^{14} s^{-1}

Explanation

Solution

vˉ=1λ=RHZ(1n12−1n22)\bar{v}=\frac{1}{\lambda}=R_{H}Z\left(\frac{1}{n^{2}_{1}}-\frac{1}{n^{2}_{2}}\right) In Balmer series n1=2&n2=3,4,5.....n_{1} = 2 \& n_{2}=3, 4, 5 ..... Last line of the spectrum is called series limit Limiting line is the line of shortest wavelength and high energy when n2=∞n_{2}=\infty ∴vˉ=1λ=RHn12=3.29×101522=3.29×10154\therefore \bar{v}=\frac{1}{\lambda}=\frac{R_{H}}{n^{2}_{1}}=\frac{3.29\times10^{15}}{2^{2}}=\frac{3.29\times10^{15}}{4} =8.22×1014s−1=8.22\times10^{14}s^{-1}