Question 1 Report
(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\(^{-1}\)].
(a) Einstein's photoelectric equation
\[ hf = W_{0} + \tfrac{1}{2}mv_{\max}^{2} \]
(b) Factors for a photocell
(c)(i) Nuclear equations
Equation A: \(^{222}_{p}\text{Rn} \to\ ^{218}_{84}\text{Po} +\ ^{q}_{2}\text{He}\). Balancing charge: \(p = 84 + 2 = 86\). Balancing mass: \(q = 222 - 218 = 4\). So \(p = 86\), \(q = 4\) (an alpha particle is emitted).
Equation B: \(^{214}_{83}\text{Bi} \to\ ^{214}_{84}\text{Po} +\ ^{m}_{n}\text{X}\). Balancing mass: \(m = 214 - 214 = 0\). Balancing charge: \(n = 83 - 84 = -1\). So \(m = 0\), \(n = -1\), and X is a beta particle (electron), \(^{0}_{-1}e\).
(ii) Radioactive waste emits ionising radiation that is harmful to living cells (it can cause cancer, genetic damage and death) and can persist for a long time, so it must be disposed of safely to protect people and the environment from radiation exposure.
(d)(i) Energy of the photon
\(E = \dfrac{hc}{\lambda} = \dfrac{6.6\times10^{-34}\times3.0\times10^{8}}{2.4\times10^{-7}} = \dfrac{19.8\times10^{-26}}{2.4\times10^{-7}} = 8.25\times10^{-19}\ \text{J}\).
This corresponds to the option \(8.2\times10^{-19}\ \text{J}\).
(ii) On the energy-level diagram, the photon of energy \(8.2\times10^{-19}\ \text{J}\) is emitted when the atom makes a downward transition (electron falling) between the two levels whose energy difference equals \(8.2\times10^{-19}\ \text{J}\); the arrow is drawn pointing downwards from the higher to the lower of those two levels. The exact levels can only be marked on the actual figure provided.
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