(a) Write Einstein's photoelectric equation and identh: each component of the equation. (b) For a...

PHYSICS
WAEC 2014

(a) Write Einstein's photoelectric equation and identh: each component of the equation.

(b) For a photocell, star; one factor each that is responsible fo: the:

(i) emission (ii) rate of emission;

(iii) energy of photoelectrons.

(c)(i) Two nuclear equations are given below:

\(^{222}_{p}RN\) \(\to\) \(^{218}_{84}PO + ^q_2He\)...................A

\(^{214}_{83}RN\) \(\to\) \(^{214}_{84}PO + ^m_nX\)...................B

Determine the values of: (\(\alpha\)) p and q in equation A; (\(\beta\)) in and n in equation B and identify X.

(ii) Give a reason why it is important to dispose o radioactive waste safely.

(d)(i) A certain atom emits ultra violet photon of wavelength 2.4 x10\(^{-7}\)m. Calculate the energy of the photon:

----------------- - 6.0 x 10\(^{-19}\)J

---------------- - 8.2 x 10\(^{-19}\)J

---------------- - 8.8 x 10\(^{-19}\)J

---------------- - 16.7 x 10\(^{-19}\)J

(ii) The figure above illustrates the energy levels o the atom. Copy the figure in your answer booklet anc indicate on it, the energy level transitions which cause the emission of the photon in (d)(i) above. [h= 6.6 x 10\(^{-34}\) Js; c = 3.0 x 108 ms\(^{-1}\)].

Explanation

(a) Einstein's photoelectric equation is

E = W + K Emax, where

E = Energy of incident radiation

W = Work function

K.E\(_{\text{max}}\) = Maximum kinetic energy of photoelectron.

(b) For photocell factors responsible for:

(i) Emission \(\to\) frequency or wavelength of incident radiation.

(ii) Rate of emission \(\to\) intensity of incident radiation or number of photons.

(iii) Energy of photoelectrons ---> frequency or wavelength of incident radiation.

(c)(i) (\(\alpha\)) 218 + q = 222

q = 4

(\(\alpha\)) P = 84 + 2 = 86,

m = 214 — 214 = 0

n = 83 — 84 = —1

x is an electron or beta particle.

ii) Needs to safely dispose radioactive waste:

(1) it emits harmful radiations.

(2) it is toxic

(d)(i) E = \(\frac{hc}{\lambda}\)

= \(\frac{6.6 \times 10^{-34} \times 3.0 \times 10^{8}}{2.4 \times 10^{8}}\)

= 8.25 x 10\(^{-19}\)J

(ii)

E\(_4\) ----------------- - 6.0 x 10\(^{-19}\)J

E\(_3\) ---------------- - 8.2 x 10\(^{-19}\)J

E\(_2\) ---------------- - 8.8 x 10\(^{-19}\)J

E\(_1\) ---------------- - 16.7 x 10\(^{-19}\)J

The energy level transitions which cause the emission of the photon in (d)(i) is. E\(_3\)



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