Question

Photoelectric work function of a metal is 1 eV. Light of wavelength $? = 3000\, \mathring{A}$ falls on it. The photo electrons come out with a maximum velocity

• A. 10 metres/sec
• B. $10^2$ metres/sec
• C. $10^4$ metres/sec
• D. $10^6$ metres/sec

Answer 1

Answer: (D)

$10^6$ metres/sec

Answer 2

$h \upsilon = W+\frac{1}{2}mv^2$ or $\frac{hc}{?}=W + \frac{1}{2}mv^2$ Here $? = 3000\, \mathring{A} = 3000 \times 10^{-10}\, m$ and $W = 1\, eV = 1.6 \times 10^{-19}$ joule $\therefore\, \, \frac{(6.6 \times 10^{-34})(3 \times10^8)}{3000 \times 10^{-10}}$ $= (6.6 \times 10^{-19})+\frac{1}{2} \times(9.1 \times10^{-31})v^2$ Solving we get $v \cong 10^6\, m/s$