When electromagnetic radiation of wavelength $300 ext{ nm}$ falls on the surface of a metal, electrons are e

Question

When electromagnetic radiation of wavelength $300 	ext{ nm}$ falls on the surface of a metal, electrons are emitted with the kinetic energy of $1.68 	imes 10^5 	ext{ J mol}^{-1}$. The minimum energy needed to remove one mole of electron from the metal is: (Given: $h = 6.626 	imes 10^{-34} 	ext{ J s}$, $c = 3 	imes 10^8 	ext{ m s}^{-1}$, $N_A = 6.022 	imes 10^{23} 	ext{ mol}^{-1}$)

Accepted Answer

1.  **Calculate the energy of one mole of photons ($E$):**
    The energy of a photon is given by $E = \frac{hc}{\lambda}$. For one mole, we multiply by Avogadro's constant ($N_A$).
    $$E = \frac{N_A hc}{\lambda}$$
    Given:
    *   $N_A = 6.022 	imes 10^{23} 	ext{ mol}^{-1}$
    *   $h = 6.626 	imes 10^{-34} 	ext{ J s}$
    *   $c = 3.0 	imes 10^8 	ext{ m s}^{-1}$
    *   $\lambda = 300 	ext{ nm} = 300 	imes 10^{-9} 	ext{ m} = 3.0 	imes 10^{-7} 	ext{ m}$

$$E = \frac{(6.022 	imes 10^{23} 	ext{ mol}^{-1})(6.626 	imes 10^{-34} 	ext{ J s})(3.0 	imes 10^8 	ext{ m s}^{-1})}{3.0 	imes 10^{-7} 	ext{ m}}$$
    $$E = \frac{11.969 	imes 10^{-3}}{3.0 	imes 10^{-7}} 	ext{ J mol}^{-1} \approx 3.99 	imes 10^5 	ext{ J mol}^{-1}$$

2.  **Apply the Photoelectric Equation:**
    Energy of photon = Minimum energy to remove electron (Work Function) + Kinetic Energy
    $$E = W_0 + K.E.$$
    $$W_0 = E - K.E.$$

3.  **Calculate the Work Function ($W_0$):**
    Given $K.E. = 1.68 	imes 10^5 	ext{ J mol}^{-1}$
    $$W_0 = (3.99 	imes 10^5 	ext{ J mol}^{-1}) - (1.68 	imes 10^5 	ext{ J mol}^{-1})$$
    $$W_0 = (3.99 - 1.68) 	imes 10^5 	ext{ J mol}^{-1}$$
    $$W_0 = 2.31 	imes 10^5 	ext{ J mol}^{-1}$$

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