Q.44. The ratio of momentum of the photons of the 1st and 2nd line of Balmer series of Hydrogen atom

Q: What is the correct answer for JEE Main Physics Q44?

The correct answer is D. Momentum of photon $p = h/\lambda$. For Balmer series, $n_f = 2$. The 1st line corresponds to $n_i = 3$, and the 2nd line corresponds to $n_i = 4$. The Rydberg formula is $\frac{1}{\lambda} = RZ^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2})$. For Hydrogen, $Z=1$. For 1st line ($n_i=3, n_f=2$): $\frac{1}{\lambda_1} = R (\frac{1}{2^2} - \frac{1}{3^2}) = R (\frac{1}{4} - \frac{1}{9}) = R (\frac{9-4}{36}) = \

Q: Q.44. The ratio of momentum of the photons of the 1st and 2nd line of Balmer series of Hydrogen atoms is . The possible values of and are :-

The correct answer is D. Momentum of photon $p = h/\lambda$. For Balmer series, $n_f = 2$. The 1st line corresponds to $n_i = 3$, and the 2nd line corresponds to $n_i = 4$. The Rydberg formula is $\frac{1}{\lambda} = RZ^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2})$. For Hydrogen, $Z=1$. For 1st line ($n_i=3, n_f=2$): $\frac{1}{\lambda_1} = R (\frac{1}{2^2} - \frac{1}{3^2}) = R (\frac{1}{4} - \frac{1}{9}) = R (\frac{9-4}{36}) = \

Q: What are the options for this JEE Main Physics question?

A. 27 and 20 · B. 3 and 16 · C. 5 and 36 · D. 20 and 27

Question

Q.44. The ratio of momentum of the photons of the 1st and 2nd line of Balmer series of Hydrogen atoms is . The possible values of  and  are :-

FirstInTest · Accepted Answer

Momentum of photon $p = h/\lambda$. For Balmer series, $n_f = 2$. The 1st line corresponds to $n_i = 3$, and the 2nd line corresponds to $n_i = 4$. The Rydberg formula is $\frac{1}{\lambda} = RZ^2 (\frac{1}{n_f^2} - \frac{1}{n_i^2})$. For Hydrogen, $Z=1$. For 1st line ($n_i=3, n_f=2$): $\frac{1}{\lambda_1} = R (\frac{1}{2^2} - \frac{1}{3^2}) = R (\frac{1}{4} - \frac{1}{9}) = R (\frac{9-4}{36}) = \frac{5R}{36}$. For 2nd line ($n_i=4, n_f=2$): $\frac{1}{\lambda_2} = R (\frac{1}{2^2} - \frac{1}{4^2}) = R (\frac{1}{4} - \frac{1}{16}) = R (\frac{4-1}{16}) = \frac{3R}{16}$. The ratio of momenta is $\frac{p_1}{p_2} = \frac{h/\lambda_1}{h/\lambda_2} = \frac{\lambda_2}{\lambda_1} = \frac{1/\lambda_1}{1/\lambda_2} = \frac{5R/36}{3R/16} = \frac{5}{36} 	imes \frac{16}{3} = \frac{5 	imes 4}{9 	imes 3} = \frac{20}{27}$. So, $\alpha/\beta = 20/27$. Possible values are $\alpha=20, \beta=27$.

Answer

27 and 20

Answer

3 and 16

Answer

5 and 36

Answer

20 and 27

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