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

If a photon of wavelength $150 \, pm$ strikes an atom and one of its inner bound electrons is ejected with a velocity of $1.5 \times 10^{7} \, ms^{-1}$, calculate the energy with which it is bound to the nucleus.

The wavelength of $H_{\alpha}$ line of Balmer series is $X \ \mathring{A}$. What is the wavelength of $H_{\beta}$ line of Balmer series?

The energy of an electron in the first Bohr orbit of $H$ atom is $-13.6 \ eV$. The energy value of an electron in the first excited state of $Li^{2+}$ is ............... $eV$.

The ratio between the radii of $He^{+}$ ion and $H$ atom is:

The electron in a hydrogen atom undergoes a transition from higher orbits to an orbit of radius $476.1 \text{ pm}$. This transition corresponds to which of the following series?

Vedclass Test Series

Exam Paper Generator

Online Exam Module