A monochromatic radiation of wavelength lambd | vedclass

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

Question

(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula: $\frac{n(

Vedclass · Accepted Answer

$97.4$

Vedclass · Answer

(C) The number of spectral lines emitted when an electron transitions from an excited state $n$ to the ground state is given by the formula: $\frac{n(n-1)}{2} = 6$.Solving for $n$: $n^2 - n - 12 = 0$, which factors to $(n-4)(n+3) = 0$. Since $n$ must be positive, $n = 4$.This means the hydrogen atoms are excited to the $n = 4$ energy level.The energy required to excite an electron from the ground state $(n_1 = 1)$ to the $n_2 = 4$ state is given by: $\Delta E = E_4 - E_1 = -13.6 \left( \frac{1}{4^2} - \frac{1}{1^2} ight) 	ext{ eV}$.$\Delta E = -13.6 \left( \frac{1}{16} - 1 ight) = -13.6 \left( -\frac{15}{16} ight) = 12.75 	ext{ eV}$.The wavelength $\lambda$ of the incident radiation is calculated using $\lambda = \frac{hc}{\Delta E}$.Given $hc = 1242 	ext{ eV-nm}$, we have $\lambda = \frac{1242}{12.75} \approx 97.4 	ext{ nm}$.

$A$ monochromatic radiation of wavelength $\lambda$ is incident on a hydrogen sample in ground state. The sample subsequently emits radiation of six different wavelengths, then the value of $\lambda$ is. [Use $hc = 1242 \text{ eV-nm}$] (in $\text{ nm}$)

Similar Questions

In a hydrogen atom,the electron is moving around the nucleus with a velocity of $2.18 \times 10^{6} \; m/s$ in an orbit of radius $0.528 \; \mathring{A}$. The acceleration of the electron is:

The ratio of the kinetic energy to the total energy of an electron in a Bohr orbit is

Which of the following graphs in the figure shows the speed $(v)$ of an electron in a hydrogen atom as a function of the principal quantum number $(n)$?

$A$ hydrogen atom is in its $n^{\text{th}}$ energy state. If the de-Broglie wavelength of the electron is $\lambda$,then:

Assertion $(A)$: The magnetic moment $(\mu)$ of an electron revolving around the nucleus decreases with increasing principal quantum number $(n)$.
Reason $(R)$: Magnetic moment of the revolving electron,$\mu \propto n$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module