Question 12 of 62intermediate🔧 ApplyNumerical2 marks

Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800 Å. Calculate threshold frequency (ν₀) and work function (W₀) of the metal.

Correct Answer

Threshold frequency (ν0\nu_0) = 4.41×10144.41 \times 10^{14} s−1^{-1} and Work function (W0W_0) = 2.92×10−192.92 \times 10^{-19} J

Exercise: EXERCISES | Q: 2.12 | (Page 41)
For More Understanding

Explanation

Since the electrons are emitted with zero velocity, the kinetic energy of the ejected electrons is zero. This means the energy of the incident photon is exactly equal to the work function of the metal (hν=hν0=W0h\nu = h\nu_0 = W_0). The given wavelength is the threshold wavelength.

Solution Steps

  1. Step 1: Convert wavelength to meters Given wavelength, λ=6800\lambda = 6800 Å =6800×10−10= 6800 \times 10^{-10} m =6.8×10−7= 6.8 \times 10^{-7} m.

  2. Step 2: Calculate threshold frequency (ν0\nu_0) Since the electrons are emitted with zero velocity, the incident radiation is at the threshold frequency. ν0=cλ=3.0×108 m s−16.8×10−7 m=4.41×1014\nu_0 = \frac{c}{\lambda} = \frac{3.0 \times 10^8 \text{ m s}^{-1}}{6.8 \times 10^{-7} \text{ m}} = 4.41 \times 10^{14} s−1^{-1}.

  3. Step 3: Calculate work function (W0W_0) The work function is the minimum energy required to eject the electron, which is W0=hν0W_0 = h\nu_0. W0=6.626×10−34 J s×4.41×1014 s−1=2.92×10−19W_0 = 6.626 \times 10^{-34} \text{ J s} \times 4.41 \times 10^{14} \text{ s}^{-1} = 2.92 \times 10^{-19} J.

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