Question 48 of 62intermediate🔧 ApplyNumerical3 marks

The work function for caesium atom is 1.9 eV. Calculate (a) the threshold wavelength and (b) the threshold frequency of the radiation. If the caesium element is irradiated with a wavelength 500 nm, calculate the kinetic energy and the velocity of the ejected photoelectron.

(a)
Answer

Threshold wavelength = 6.53×10−76.53 \times 10^{-7} m (or 653 nm)

(b)
Answer

Threshold frequency = 4.59×10144.59 \times 10^{14} s−1^{-1}

Explanation

To solve this problem, we use the photoelectric effect equations. The work function (W0W_0) is the minimum energy required to eject an electron, which relates to threshold frequency (ν0\nu_0) and threshold wavelength (λ0\lambda_0). When radiation of a specific wavelength strikes the metal, the excess energy after overcoming the work function becomes the kinetic energy of the ejected electron. Using the kinetic energy, we can calculate the velocity of the electron.

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