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Question: Binding energy of electrons in a metal is 250 KJ $mol^{-1}$. Then threshold frequency of metal is...

Binding energy of electrons in a metal is 250 KJ mol−1mol^{-1}. Then threshold frequency of metal is

A

ID: 2025075891 3.8x1038s−13.8x10^{38}s^{-1}

B

ID: 2025075891 3.8x1035s−13.8x10^{35}s^{-1}

C

ID: 2025075891 6.3x1011s−16.3x10^{11}s^{-1}

D

ID: 2025075891 6.3x1014s6.3x10^{14}s

Answer

D

Explanation

Solution

The binding energy of electrons in a metal is given as 250 KJ mol−1mol^{-1}. This binding energy is equivalent to the work function (Φ\Phi) per mole. To find the work function for a single electron, we need to divide the molar binding energy by Avogadro's number (NAN_A).

  1. Convert molar binding energy to Joules per mole: Binding energy = 250 KJ mol−1=250×103 J mol−1250 \text{ KJ mol}^{-1} = 250 \times 10^3 \text{ J mol}^{-1}

  2. Calculate the work function (Φ\Phi) per electron: The work function (Φ\Phi) is the minimum energy required to eject one electron. Φ=Binding energy per moleNA\Phi = \frac{\text{Binding energy per mole}}{N_A} Using Avogadro's number (NA=6.022×1023 mol−1N_A = 6.022 \times 10^{23} \text{ mol}^{-1}): Φ=250×103 J mol−16.022×1023 mol−1\Phi = \frac{250 \times 10^3 \text{ J mol}^{-1}}{6.022 \times 10^{23} \text{ mol}^{-1}} Φ=2.5×1056.022×1023 J\Phi = \frac{2.5 \times 10^5}{6.022 \times 10^{23}} \text{ J} Φ≈4.151×10−19 J\Phi \approx 4.151 \times 10^{-19} \text{ J}

  3. Calculate the threshold frequency (ν0\nu_0): The work function is related to the threshold frequency by Planck's equation: Φ=hν0\Phi = h\nu_0 Where hh is Planck's constant (h=6.626×10−34 J sh = 6.626 \times 10^{-34} \text{ J s}). Rearranging the formula to find ν0\nu_0: ν0=Φh\nu_0 = \frac{\Phi}{h} ν0=4.151×10−19 J6.626×10−34 J s\nu_0 = \frac{4.151 \times 10^{-19} \text{ J}}{6.626 \times 10^{-34} \text{ J s}} ν0≈0.6265×1015 s−1\nu_0 \approx 0.6265 \times 10^{15} \text{ s}^{-1} ν0≈6.265×1014 s−1\nu_0 \approx 6.265 \times 10^{14} \text{ s}^{-1}

  4. Compare with given options: Rounding the calculated value to one decimal place, we get 6.3×1014 s−16.3 \times 10^{14} \text{ s}^{-1}. Option D is 6.3×1014 s6.3 \times 10^{14} \text{ s}. Although the unit is incorrectly given as 's' (which is for period) instead of 's−1^{-1}' (which is for frequency), the numerical value matches our calculation. It is a common occurrence in multiple-choice questions for there to be a typo in the unit, but the numerical value is intended to be correct.