The electric field at the point associated wi | vedclass

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

Question

(D) The electric field is given by the sum of two waves with angular frequencies $\omega_1 = 6 	imes 10^{15} \, rad/s$ and $\omega_2 = 9 	imes 10^{1

Vedclass · Accepted Answer

$3.42$

Vedclass · Answer

(D) The electric field is given by the sum of two waves with angular frequencies $\omega_1 = 6 	imes 10^{15} \, rad/s$ and $\omega_2 = 9 	imes 10^{15} \, rad/s$.To find the maximum kinetic energy,we consider the photon with the higher frequency,as $K_{max} = hf - \phi$.The frequency $f$ is given by $f = \frac{\omega}{2\pi}$.For the higher frequency component,$f = \frac{9 	imes 10^{15}}{2\pi} \, Hz$.The energy of the photon is $E_{photon} = hf = (4.14 	imes 10^{-15} \, eVs) 	imes \left( \frac{9 	imes 10^{15}}{2 	imes 3.14159} ight) \, Hz$.$E_{photon} = \frac{37.26}{6.283} \approx 5.93 \, eV$.The maximum kinetic energy is $K_{max} = E_{photon} - \phi = 5.93 \, eV - 2.50 \, eV = 3.43 \, eV$.The closest option provided is $3.42 \, eV$.

Similar Questions

In a photoemissive cell with exciting wavelength $\lambda$,the fastest electron has speed $v$. If the exciting wavelength is changed to $\frac{3\lambda}{4}$,the speed of the fastest emitted electron will be

$A$ photon of energy $8 eV$ is incident on a metal surface of threshold frequency $1.6 \times 10^{15} Hz$. The maximum kinetic energy of the emitted photoelectrons is ............... $eV$ (Given $h = 6.6 \times 10^{-34} Js$).

The maximum velocity of emitted photoelectrons is $4.8 \ m/s$. If the $e/m$ ratio of an electron is $1.76 \times 10^{11} \ C/kg$,then the stopping potential is ......

When light of intensity $1 \ W/m^2$ and wavelength $5 \times 10^{-7} \ m$ is incident on a surface, it is completely absorbed. If $100$ photons emit one electron and the surface area is $1 \ cm^2$, what is the photoelectric current?

Vedclass Test Series

Exam Paper Generator

Online Exam Module