What is the work function of the metal if the | vedclass

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

Question

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{4000 	imes 10^{-10}

Vedclass · Accepted Answer

$2.1$

Vedclass · Answer

(C) The energy of the incident photon is given by $E = \frac{hc}{\lambda}$.$E = \frac{6.626 	imes 10^{-34} 	imes 3 	imes 10^8}{4000 	imes 10^{-10}} = 4.9695 	imes 10^{-19} \, J$.The kinetic energy of the emitted photoelectron is $K.E. = \frac{1}{2}mv^2$.$K.E. = \frac{1}{2} 	imes 9 	imes 10^{-31} 	imes (6 	imes 10^5)^2 = 1.62 	imes 10^{-19} \, J$.According to Einstein's photoelectric equation,$E = \phi + K.E.$,where $\phi$ is the work function.$\phi = E - K.E. = 4.9695 	imes 10^{-19} - 1.62 	imes 10^{-19} = 3.3495 	imes 10^{-19} \, J$.To convert the work function into $eV$,divide by the charge of an electron $(1.6 	imes 10^{-19} \, C)$:$\phi = \frac{3.3495 	imes 10^{-19}}{1.6 	imes 10^{-19}} \approx 2.093 \, eV \approx 2.1 \, eV$.

Similar Questions

The correct representation of the wavelength-intensity relationship of an ideal black body radiation at two different temperatures $T_{1}$ and $T_{2}$ (where $T_{2} > T_{1}$) is:

The wave numbers of three spectral lines of $H$ atom are considered. Identify the set of spectral lines belonging to the Balmer series. ($R =$ Rydberg constant)

The frequency of radiation absorbed or emitted when transition occurs between two stationary states with energies $E_1$ (lower) and $E_2$ (higher) is given by

The radius of the second orbit of a hydrogen atom is the same as that of orbit $n$ of an ion $x$. $n$ and $x$ are respectively:

The electron in the $n^{th}$ orbit of $Li^{2+}$ is excited to $(n+1)$ orbit using the radiation of energy $1.47 \times 10^{-17} \ J$. The value of $n$ is $....$. Given $R_H = 2.18 \times 10^{-18} \ J$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module