Question 18M.2.HL.TZ1.6

${}_4^{10}\text{Be}{\to }_5^{10}\text{B}{+}_{-1}^0\text{e}+{\overline{\text{V}}}_{\text{e}}$

antineutrino AND charge AND mass number of electron ${}_{-1}^0\text{e}$, $\overline{\text{V}}$

conservation of mass number AND charge ${}_5^{10}\text{B}$, ${}_4^{10}\text{Be}$

Do not accept V.

Accept $\overline{V}$ without subscript e.

[2 marks]

correct shape ie increasing from 0 to about 0.80 N₀

crosses given line at 0.50 N₀

[2 marks]

ALTERNATIVE 1

fraction of Be = $\frac{1}{8}$, 12.5%, or 0.125

therefore 3 half lives have elapsed

$t_{\frac{1}{2}}=\frac{4.3\times {10}^6}{3}=1.43\times {10}^6$ «≈ 1.4 × 10⁶» «y»

ALTERNATIVE 2

fraction of Be = $\frac{1}{8}$, 12.5%, or 0.125

$\frac{1}{8}={\text{e}}^{-\lambda }(4.3\times {10}^6)$ leading to λ = 4.836 × 10^–7 «y»^–1

$\frac{\ln 2}{\lambda }$ = 1.43 × 10⁶ «y»

Must see at least one extra sig fig in final answer.

[3 marks]

λ «= $\frac{\ln 2}{1.4\times {10}^6}$» = 4.95 × 10^–7 «y^–1»

rearranging of A = λN₀e^–λt to give –λt = ln $\frac{8.0\times {10}^{-3}\times 365\times 24\times 60\times 60}{4.95\times {10}^{-7}\times 7.6\times {10}^{11}}$ «= –0.400»

t = $\frac{-0.400}{-4.95\times {10}^{-7}}=8.1\times {10}^5$ «y»

Allow ECF from MP1

[3 marks]

emission of (infrared) electromagnetic/infrared energy/waves/radiation.

[1 mark]

the (peak) wavelength of emitted em waves depends on temperature of emitter/reference to Wein’s Law

so frequency/color depends on temperature

[2 marks]

$\lambda =\frac{2.90\times {10}^{-3}}{253}$

= 1.1 × 10^–5 «m»

Allow ECF from MP1 (incorrect temperature).

[2 marks]

from the laboratory to the sample

conduction – contact between ice and lab surface.

OR

convection – movement of air currents

Must clearly see direction of energy transfer for MP1.

Must see more than just words “conduction” or “convection” for MP2.

[2 marks]