Ncert Exercises & Solutions

Question 11.8:
The threshold frequency for a certain metal is 3.3 x $10^{14}$ Hz. If the light of frequency 8.2x10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Hint: Use Einstein's photoelectric equation.
Step 1: Find the cut-off voltage.

Threshold frequency of the metal, $\nu_{0}$ = 3.3 x $10^{14}$ Hz
Frequency of light incident on the metal, $\nu$ = 8.2 x $10^{14}$ Hz
Charge on an electron, e = 1.6 x $10^{- 19}$ C
Planck's constant, h = 6.626 x $10^{- 34}$ Js
Cut-off voltage for the photoelectric emission from the metal = $V_{0}$
The equation for the cut-off energy is given as:
$\begin{array}{l} e V_{0} = h (v - v_{0}) \\ V_{0} = \frac{h (v - v_{0})}{e} \\ = \frac{6.626 \times 10^{- 34} \times (8.2 \times 10^{14} - 3.3 \times 10^{14})}{1.6 \times 10^{- 19}} = 2.0292 V. \end{array}$

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