Explanation

(a) electrons are only emitted when the photon frequency exceeds the threshold frequency,

- the energy of each photon is proportional to its frequency,

- the emission of photoelectrons depends on the frequency of incident light on the metal,

- the intensity of light does not affect the energy of the electrons emitted/photoelectrons,

- the energy of the photoelectron depends on the frequency of incident radiation.

(b) Given: P = 6 x 10\(^{-3}\) W, no of photons per seconds = 1 x 10\(^{16}\), m\(_e\) = 9.1 x 10\(^{-31}\) kg, e = 1.6 x 10\(^{-19}\) C, W\(_o\) = ?

Photon Energy E = \(\frac{\text{power}}{\text{no of photons per second}}\) = \(\frac{\text{P}}{\text{n}}\)

= \(\frac{6.0 \times 10^{-3}}{1 \times 10^{16}}\)

 =  6 x 10\(^{-19}\)J

But K.E = E +  W\(_o\)

W\(_o\) = E - K.E = E - eV

W\(_o\) = 6 x 10\(^{-19}\) - 2.2 x (1.6 x 10\(^{-19}\)) = 2.48 x 10\(^{-9}\)J

(c) Factors on which radioactive activity depends are: (i) decay constant, (ii) number of unstable nuclei present at a given time (iii) time of the decay process (iv) environmental conditions, (v) type of radioisotope, etc

(d) Given: A = 2.0 x 10\(^6\)Bq, t\(_{\frac{1}{2}}\) = 1.8 x 10\(^8\)s

A = \(\lambda\)N, where \(\lambda\) = decay constant, N = number of radioactive nuclei, A = activity.

But  \(\lambda\) = \(\frac{In(2)}{t_{\frac{1}{2}}}\)

 \(\lambda\) = \(\frac{In(2)}{1.8 \times 10^8}\) ≈ \(\frac{0.693}{1.8 \times 10^8}\) ≈ 3.85 x 10\(^{-9}\)s\(^{-1}\)

Recall, A = \(\lambda\)N

N = \(\frac{A}{\lambda}\) = \(\frac{2.0 \times 10^6}{3.85 \times 10^{-9}}\) = 5.2 x 10\(^{14}\) nuclei

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