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Ultraviolet radiations of 6.2 eV falls on an aluminium surface. K.E. of fastest electron emitted is (work function = 4.2 eV):

1. 3.2 × 10⁻²¹ J
2. 3.2 × 10⁻¹⁹ J
3. 7 × 10⁻¹⁹ J
4. 9 × 10⁻³² J

Correct answer: 3.2 × 10⁻¹⁹ J

Solution

The kinetic energy of the fastest electron is given by Einstein's photoelectric equation: K.E. = hν - φ, where hν is the energy of the incident photon and φ is the work function. Substituting the given values, K.E. = 6.2 eV - 4.2 eV = 2.0 eV. Converting to joules (1 eV = 1.6 × 10⁻¹⁹ J), K.E. = 2 × 1.6 × 10⁻¹⁹ J = 3.2 × 10⁻¹⁹ J.

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