Electrons are emitted with zero velocity from | vedclass

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

Question

(B) The threshold wavelength $\lambda_0$ is $8000 \ \mathring{A} = 8000 	imes 10^{-10} \ m$. The work function $\Phi = \frac{hc}{\lambda_0} = \frac{6

Vedclass · Accepted Answer

$5.74 	imes 10^5 \ ms^{-1}$

Vedclass · Answer

(B) The threshold wavelength $\lambda_0$ is $8000 \ \mathring{A} = 8000 	imes 10^{-10} \ m$. The work function $\Phi = \frac{hc}{\lambda_0} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{8000 	imes 10^{-10}} = 2.475 	imes 10^{-19} \ J$. When exposed to $\lambda = 5000 \ \mathring{A} = 5000 	imes 10^{-10} \ m$,the energy of incident radiation is $E = \frac{hc}{\lambda} = \frac{6.6 	imes 10^{-34} 	imes 3 	imes 10^8}{5000 	imes 10^{-10}} = 3.96 	imes 10^{-19} \ J$. The kinetic energy $K.E. = E - \Phi = 3.96 	imes 10^{-19} - 2.475 	imes 10^{-19} = 1.485 	imes 10^{-19} \ J$. Using $K.E. = \frac{1}{2} m_e v^2$,we get $v = \sqrt{\frac{2 	imes K.E.}{m_e}} = \sqrt{\frac{2 	imes 1.485 	imes 10^{-19}}{9.1 	imes 10^{-31}}} = \sqrt{0.326 	imes 10^{12}} \approx 5.71 	imes 10^5 \ ms^{-1}$. Rounding to the nearest option,the answer is $5.74 	imes 10^5 \ ms^{-1}$.

Similar Questions

Find the energy of radiation having $7000 \ \mathring{A}$ and $400 \ nm$ wavelength. $\left( h = 6.62 \times 10^{-34} \ J \ s, c = 3 \times 10^8 \ m/s \right)$

Splitting of spectral lines under the influence of a magnetic field is called:

The wavelength (in $\mathring{A}$) of a photon having energy $3 \ eV$ is approximately.
$1 \ eV = 1.6 \times 10^{-12} \ erg$
$h = 6.626 \times 10^{-27} \ erg \ s$
$c = 3 \times 10^{10} \ cm/s$

Calculate $(a)$ wavenumber and $(b)$ frequency of yellow radiation having wavelength $5800 \, \mathring{A}$.

Which of the following radiations has the highest energy?

Vedclass Test Series

Exam Paper Generator

Online Exam Module