Question

The threshold frequency for a certain metal is 3.3 X 10¹⁴ Hz. If light of frequency 8.2 X 10¹⁴ Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission.

Answer 1

Answer:

The cut-off voltage for the photoelectric emission = 2.0292 Volt.

Explanation:

As per the question,

Given data are:

The threshold frequency for a certain metal, v₀= 3.3 × 10¹⁴ Hz

Frequency, v = 8.2 × 10¹⁴ Hz

As we know,

Charge of an electron, e = 1.6 × 10⁻¹⁹C

Planck's constant, h = 6.63 × 10⁻³⁴Js

According to the equation of cut-off energy, we have:

eV₀ = h ( v - v₀ )

Where,

V₀ = cut-off voltage for the photoelectric emission

Now,

Put all the given data in this equation, we get:

eV₀ = h ( v - v₀ )

⇒ 1.6 × 10⁻¹⁹ V₀ = 6.63 × 10⁻³⁴ ( 8.2 × 10¹⁴ - 3.3 × 10¹⁴ )

⇒ V₀ = 2.0292 Volt

Hence, the cut-off voltage for the photoelectric emission = 2.0292 Volt.

Answer 2

Solution :-

We know the photoelectric effect equation

$eV_a=K.E_c-K.E_e$

where

$e=Charge\ on\ an\ electron\\V_a=Variable\ voltage\\K.E_c=Kinetic\ energy\ of\ electrons\ at\ collector\ plate\\K.E_e=Kinetic\ energy\ of\ electron\ at\ emitter\ plate$

At cutoff voltage, even, higher energy electrons are unable to reach at collector plate and polarity of original battery is reversed to produce an Electric field opposing motion of electrons.

So, At cutoff voltage,

$V_a=-V_s, \ and\ K.E_e=K.E_{max};\ K.E_c=0$

So,$=>-eV_s=-K.E_{max}\\=>eV_s=hf-\phi,( \ where\ h=Planck's \ constant;f=frequency\ of\ emitted\ light;\phi=work\ function)\\=>eV_s=hf-hf_o,(\ where\ f_o=Threshold\ frequency)\\=>V_s=\frac{h}{e} (f-f_o)\\=>V_s=\frac{h}{e} (8.2\times10^{14}-3.3\times10^{14})\\=>V_s=\frac{6.626\times10^{-34}\times4.9\times10^{14}}{1.6\times10^{19}} \\=>V_s=2$Hence, required cutoff voltage would be 2 Volts.

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