Question

Photoelectric work function of a metal 1 eV. If a light of wavelength 500 Å falls on the metal, the speed of the photoelectron emitted approximately will be- ( h = 6.625 × 10^–34 J s)

• A. Equal to velocity light
• B. $\frac{1}{10}$× velocity of light
• C. $\frac{1}{100}$× velocity of light
• D. $\frac{1}{1000}$× velocity of light

Answer 1

Answer: (C)

$\frac{1}{100}$× velocity of light

Answer 2

hn = w + KE = w + $\frac{1}{2}$m_e v²

or h $\frac{c}{\lambda}$= 1 + $\frac{1}{2}$ m_e v²

or $\frac{6.625 \times 10^{–34} \times 3 \times 10^{8}}{500 \times 10^{–10}}$= 1.6 × 10^–19 + $\frac{1}{2}$ × 9.1 × 10^–31 × v²

or $\frac{1}{2}$× 9.1 × 10^–31 × v²

= $\frac{6.625 \times 3}{5}$ × 10^–18 –1.6 × 10^–19

= 38.15 × 10^–19

or v² = $\frac{38.15 \times 2 \times 10^{12}}{9.1}$

or v² = 8.385 × 10¹²

or v = 2. 89 × 10⁶ m/s

\ ­ $\frac{v}{c}$ ~ $\frac{3 \times 10^{6}}{3 \times 10^{8}}$= $\frac{1}{100}$