Energy required to remove an electron from an | vedclass

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

Question

(A) The energy of the incident photon is given by $E = \frac{12400}{\lambda (	ext{\AA})} 	ext{ eV} = \frac{12400}{2000} 	ext{ eV} = 6.2 	ext{ eV}$

Vedclass · Accepted Answer

$8.4 	imes 10^5 	ext{ m/s}$

Vedclass · Answer

(A) The energy of the incident photon is given by $E = \frac{12400}{\lambda (	ext{\AA})} 	ext{ eV} = \frac{12400}{2000} 	ext{ eV} = 6.2 	ext{ eV}$.Using Einstein's photoelectric equation: $E = W_0 + K_{\max}$, where $W_0 = 4.2 	ext{ eV}$ is the work function.$K_{\max} = E - W_0 = 6.2 	ext{ eV} - 4.2 	ext{ eV} = 2.0 	ext{ eV}$.Converting to Joules: $K_{\max} = 2.0 	imes 1.6 	imes 10^{-19} 	ext{ J} = 3.2 	imes 10^{-19} 	ext{ J}$.Since $K_{\max} = \frac{1}{2}mv_{\max}^2$, we have $v_{\max} = \sqrt{\frac{2K_{\max}}{m}}$.Substituting $m = 9.1 	imes 10^{-31} 	ext{ kg}$:$v_{\max} = \sqrt{\frac{2 	imes 3.2 	imes 10^{-19}}{9.1 	imes 10^{-31}}} = \sqrt{0.703 	imes 10^{12}} \approx 0.838 	imes 10^6 	ext{ m/s} \approx 8.4 	imes 10^5 	ext{ m/s}$.

Energy required to remove an electron from an aluminium surface is $4.2 \text{ eV}$. If light of wavelength $2000 \text{ \AA}$ falls on the surface, the velocity of the fastest electron ejected from the surface will be:

Similar Questions

The ratio of the ionization energy of a Bohr hydrogen atom to that of a Bohr hydrogen-like lithium atom is:

$A$ block of mass $2 ~kg$ is initially at rest on a horizontal frictionless surface. $A$ horizontal force $\overrightarrow{F} = (9 - x^2) \hat{i} ~N$ acts on it,starting from $x = 0$. The maximum kinetic energy of the block between $x = 0$ and $x = 3 ~m$ in joules is:

In the circuit shown in the figure,the power dissipated as heat in the $6\,\Omega$ resistor is $6\,W$. What is the value of resistance $R$ in the circuit? $\Omega$

Thermodynamically the most stable form of carbon is

Vedclass Test Series

Exam Paper Generator

Online Exam Module