A hydrogen atom in the ground state is excite | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \; 	ext{eV} \cdot \mathring{A}$,we get $E = \frac{

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.Using $hc \approx 12400 \; 	ext{eV} \cdot \mathring{A}$,we get $E = \frac{12400}{975} \approx 12.75 \; 	ext{eV}$.The energy levels of a hydrogen atom are given by $E_n = -\frac{13.6}{n^2} \; 	ext{eV}$.For the ground state $(n=1)$,$E_1 = -13.6 \; 	ext{eV}$.After absorbing the photon,the new energy level is $E_n = E_1 + E = -13.6 + 12.75 = -0.85 \; 	ext{eV}$.Since $E_n = -\frac{13.6}{n^2} = -0.85 \; 	ext{eV}$,we have $n^2 = \frac{13.6}{0.85} = 16$,so $n = 4$.The electron is excited to the $n = 4$ state.The number of spectral lines emitted when the electron transitions from $n$ to the ground state is given by $\frac{n(n-1)}{2}$.For $n = 4$,the number of spectral lines is $\frac{4(4-1)}{2} = \frac{4 	imes 3}{2} = 6$.

$A$ hydrogen atom in the ground state is excited by monochromatic radiation of $\lambda = 975 \; \mathring{A}$. The number of spectral lines in the resulting emission spectrum will be:

Similar Questions

If radiation of all wavelengths from ultraviolet to infrared is passed through hydrogen gas at room temperature,absorption lines will be observed in the:

In a hydrogen atom,a transition from the state $n=6$ to $n=1$ results in ultraviolet radiation. Infrared radiation will be obtained in the transition:

The wavelength of the yellow line of sodium is $5896 \mathring{A}$. Its wave number will be:

Let the series limit for the Balmer series be $\lambda_{1}$ and the longest wavelength for the Brackett series be $\lambda_{2}$. Then $\lambda_{1}$ and $\lambda_{2}$ are related as:

Which of the following is true for the number of spectral lines when going from the Lyman series to the Pfund series?

Vedclass Test Series

Exam Paper Generator

Online Exam Module