The photon having 6000 mathring{A} wavelengt | vedclass

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

Question

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is

Vedclass · Accepted Answer

$3000 \ \mathring{A}$

Vedclass · Answer

(A) The energy of a photon is given by the formula $E = \frac{hc}{\lambda}$,where $h$ is Planck's constant,$c$ is the speed of light,and $\lambda$ is the wavelength. Since $h$ and $c$ are constants,$E \propto \frac{1}{\lambda}$. Given $E_1 = E$ and $\lambda_1 = 6000 \ \mathring{A}$. For $E_2 = 2E$,we have $\frac{E_2}{E_1} = \frac{\lambda_1}{\lambda_2}$. Substituting the values: $\frac{2E}{E} = \frac{6000 \ \mathring{A}}{\lambda_2}$. $2 = \frac{6000 \ \mathring{A}}{\lambda_2}$. $\lambda_2 = \frac{6000 \ \mathring{A}}{2} = 3000 \ \mathring{A}$.

The photon having $6000 \ \mathring{A}$ wavelength has energy $E$. If the energy of photon is $2E$,find the wavelength of the photon.

Similar Questions

The nuclear theory of the atom was put forward by:

The ratio of the radius of the first orbit in hydrogen to the radius of the first orbit in deuterium is .......

The radii of the first Bohr orbit of $H$ $(r_{H})$,$He^{+}$ $(r_{He^{+}})$ and $Li^{2+}$ $(r_{Li^{2+}})$ are in the order

If each hydrogen atom is excited by giving $8.4 \ eV$ of energy,then the number of spectral lines emitted is equal to :

The frequency of one of the lines in the Paschen series of a hydrogen atom is $2.340 \times 10^{11} \ Hz$. The quantum number $n_2$ which produces this transition is

Vedclass Test Series

Exam Paper Generator

Online Exam Module