The energy of a photon with a wavelength of 4 | 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 $E = hf = \frac{hc}{\lambda}$.Given:Planck's constant $h = 6.63 	imes 10^{-34} \, J \cdot s$Speed

Vedclass · Accepted Answer

$4.4 	imes 10^{-19} \, J$

Vedclass · Answer

(A) The energy of a photon is given by the formula $E = hf = \frac{hc}{\lambda}$.Given:Planck's constant $h = 6.63 	imes 10^{-34} \, J \cdot s$Speed of light $c = 3 	imes 10^8 \, m/s$Wavelength $\lambda = 450 \, nm = 450 	imes 10^{-9} \, m$Substituting the values:$E = \frac{(6.63 	imes 10^{-34}) 	imes (3 	imes 10^8)}{450 	imes 10^{-9}}$$E = \frac{19.89 	imes 10^{-26}}{450 	imes 10^{-9}}$$E = 0.0442 	imes 10^{-17} \, J = 4.42 	imes 10^{-19} \, J$Rounding to two significant figures,$E \approx 4.4 	imes 10^{-19} \, J$.

The energy of a photon with a wavelength of $450 \, nm$ is ............

Similar Questions

Which of the following is true for a photon?

What is the energy and momentum of a photon having frequency $v$?

The minimum intensity of light to be detected by the human eye is $10^{-10} \ W/m^2$. The number of photons of wavelength $5.6 \times 10^{-7} \ m$ entering the eye,with a pupil area of $10^{-6} \ m^2$,per second for vision will be nearly:

Write the equation for the energy of a photon.

When a photon enters glass from air,which one of the following quantities does $\underline{\text{NOT}}$ change?

Vedclass Test Series

Exam Paper Generator

Online Exam Module