Question

$K_{max} \left(\right. \text{in} \, e V \left.\right) = 3 ,$ and frequency of light $(in H z \left.\right) = 1 \times \left(10\right)^{15}$ for a metal used as a cathode in a photoelectric experiment. The threshold frequency of light for the photoelectric emission from the metal is -

• A. $1 \times 10^{14} \text{ Hz}$
• B. $1.5 \times 10^{14} \, \text{Hz}$
• C. $2.1 \times 10^{14} \text{ Hz}$
• D. $2.7 \times 10^{14} \, \text{Hz}$

Answer 1

Answer: (D)

Given, $\text{K}_{max} = 3 \, \, \text{eV}$
$h = 4.125 \times 10^{- 15} \, \text{eV} \, \text{Hz}^{\text{-1}}$
$h\nu=K_{max}+W$
$4.125\times 10^{- 15}\times 10^{15}=3+W$
or $4.125= \, 3 \, +W$
or $W=1.125$
The threshold frequency
$\nu_{0} = \frac{W}{h}$
$\nu_{0} = \frac{1.125}{4.125 \times 10^{- 15}}$
$\nu_{0} = \frac{1.125 \times 10^{- 15}}{4.125}$
$\nu_{0} = 2.72 \times 10^{14}$