Q.36. In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orb

Q: What is the correct answer for this exam Physics Q36?

The correct answer is C. Sol. Given: $\frac{r_i}{r_f} = \frac{16}{4} = 4$. For hydrogen atom, $r_n \propto n^2$. So, $\frac{n_i^2}{n_f^2} = 4 \Rightarrow \frac{n_i}{n_f} = 2$. Let $n_f = x$, then $n_i = 2x$. Since $r_f = 4$, $n_f^2 = 4 \Rightarrow n_f = 2$. Then $n_i = 2n_f = 2(2) = 4$. So the transition is from $n_i = 4$ to $n_f = 2$. The Rydberg formula for wavelength is $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \

Q: Q.36. In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f). The ratio of the radius of the orbits is given by . The wavelength of photon emitted due to…

The correct answer is C. Sol. Given: $\frac{r_i}{r_f} = \frac{16}{4} = 4$. For hydrogen atom, $r_n \propto n^2$. So, $\frac{n_i^2}{n_f^2} = 4 \Rightarrow \frac{n_i}{n_f} = 2$. Let $n_f = x$, then $n_i = 2x$. Since $r_f = 4$, $n_f^2 = 4 \Rightarrow n_f = 2$. Then $n_i = 2n_f = 2(2) = 4$. So the transition is from $n_i = 4$ to $n_f = 2$. The Rydberg formula for wavelength is $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \

Q: What are the options for this this exam Physics question?

A. 121 · B. 242 · C. 486 · D. 974

Question

Q.36. In the hydrogen atom, the electron makes a transition from the higher orbit (i) to a lower orbit (f). The ratio of the radius of the orbits is given by . The wavelength of photon emitted due to this transition is ______ nm. (Given Rydberg constant = )

FirstInTest · Accepted Answer

Sol. Given: $\frac{r_i}{r_f} = \frac{16}{4} = 4$. For hydrogen atom, $r_n \propto n^2$. So, $\frac{n_i^2}{n_f^2} = 4 \Rightarrow \frac{n_i}{n_f} = 2$. Let $n_f = x$, then $n_i = 2x$. Since $r_f = 4$, $n_f^2 = 4 \Rightarrow n_f = 2$. Then $n_i = 2n_f = 2(2) = 4$. So the transition is from $n_i = 4$ to $n_f = 2$. The Rydberg formula for wavelength is $\frac{1}{\lambda} = R \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} ight)$. $\frac{1}{\lambda} = 1.0973 	imes 10^7 \left( \frac{1}{2^2} - \frac{1}{4^2} ight) = 1.0973 	imes 10^7 \left( \frac{1}{4} - \frac{1}{16} ight) = 1.0973 	imes 10^7 \left( \frac{4-1}{16} ight) = 1.0973 	imes 10^7 	imes \frac{3}{16}$. $\lambda = \frac{16}{3 	imes 1.0973 	imes 10^7} = \frac{16}{3.2919 	imes 10^7} \approx 4.86 	imes 10^{-7} m = 486 	imes 10^{-9} m = 486 nm$.

Answer

121

Answer

242

Answer

486

Answer

974

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