When light of wavelength lambda incidents on | vedclass

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

Question

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $E$ is given by: $E = \frac{hc}{\lambda} - \phi$ ... $(i)$,where $\phi$

Vedclass · Accepted Answer

$\frac{hc}{2 \lambda}$

Vedclass · Answer

(B) According to Einstein's photoelectric equation,the maximum kinetic energy $E$ is given by: $E = \frac{hc}{\lambda} - \phi$ ... $(i)$,where $\phi$ is the work function.When the wavelength is reduced by $50 \%$,the new wavelength $\lambda' = \lambda - 0.5\lambda = 0.5\lambda = \frac{\lambda}{2}$.The new maximum kinetic energy $E' = 3E$.Substituting these into the photoelectric equation: $3E = \frac{hc}{\lambda/2} - \phi = \frac{2hc}{\lambda} - \phi$ ... $(ii)$.From equation $(i)$,we have $E = \frac{hc}{\lambda} - \phi$. Substituting this into equation $(ii)$:$3(\frac{hc}{\lambda} - \phi) = \frac{2hc}{\lambda} - \phi$$\frac{3hc}{\lambda} - 3\phi = \frac{2hc}{\lambda} - \phi$$\frac{3hc}{\lambda} - \frac{2hc}{\lambda} = 3\phi - \phi$$\frac{hc}{\lambda} = 2\phi$$\phi = \frac{hc}{2\lambda}$.

Similar Questions

Why maximum kinetic energy of a photoelectron cannot be negative?

The variation of stopping potential $V_{0}$ with frequency $\nu$ of incident radiation for a given photosensitive material is a straight line [frequency $\nu$ of incident radiation is greater than threshold frequency $\nu_{0}$]. The slope of this line is . . . . . . .

In the photoelectric effect,if the intensity of light is doubled,then the maximum kinetic energy of photoelectrons will become

Light of wavelength $\lambda$ which is less than threshold wavelength is incident on a photosensitive material. If incident wavelength is decreased so that emitted photoelectrons are moving with the same velocity,then stopping potential will:

The work function of a photosensitive material is $4.0 \ eV$. The longest wavelength of light that can cause photon emission from the substance is approximately $...... \ nm$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module