Question

The work function of a tungsten is $4.50eV$. The wavelength of fastest electron emitted when light whose photon energy is $5.50eV$ falls on tungsten surface, is:

• A. $11.73A$
• B. $0.286A$
• C. $12400A$
• D. $1.227A$

Answer 1

The correct option is A $11.73A$

Maximum kinetic energy of electron =$h\nu -\varphi$

$=(5.50-4.50)eV$

$=1eV$

$=1.6\times {10}^{-19}J$

or, $\frac{p^2}{2m}=1.6\times {10}^{-19}$

$\therefore p^2=3.2\times m\times {10}^{-19}$

$p=\sqrt{3.2\times m\times {10}^{-19}}$

$\therefore \lambda =\frac{h}{p}=\frac{6.626\times {10}^{-34}}{3.2\times {10}^{-19}\times 9.1\times {10}^{-31}}$

$=\frac{6.626\times {10}^{-34}}{5.643\times {10}^{-25}}$

$=1.173\times {10}^{-9}$

$=11.73\times {10}^{-10}m$

$=11.73A$