Answer either subparts (i) to (iii) or (iv) to (vii). Draw a labelled graph showing the variation…
Answer either subparts (i) to (iii) or (iv) to (vii).
(i)[2.0]
Draw a labelled graph showing the variation of binding energy per nucleon with the mass number (A) of the nucleus. On it, mark the nucleus that is most stable.
Draw: graph showing the variation of binding energy per nucleon with the mass number (A) of the nucleus
Must show: peak around A = 56 for Fe-56, drop at low A with peaks at He-4, C-12, O-16, gradual decrease for heavy nuclei
(ii)[1.0]
What is meant by the following statement?
'Angular momentum of an orbiting electron is quantised.'
(iii)[2.0]
Calculate the shortest wavelength of Balmer series.
Answer
Answer (ii)
AIWritten by AI - it can contain mistakes.
It means that the orbital angular momentum ($L$) of an electron revolving around the nucleus can only take discrete values that are integral multiples of $\frac{h}{2\pi}$ ($L = mvr = \frac{nh}{2\pi}$, where $n = 1, 2, 3\ldots$), and cannot vary continuously.
Answer (iii)
AIWritten by AI - it can contain mistakes.
Using the Rydberg formula for hydrogen spectrum:
$\frac{1}{\lambda} = R_{H}\left(\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}}\right)$
For the Balmer series, $n_{1} = 2$. The shortest wavelength (series limit) occurs when $n_{2} = \infty$:
$\frac{1}{\lambda_{\min}} = R_{H}\left(\frac{1}{2^{2}} - \frac{1}{\infty^{2}}\right) = \frac{R_{H}}{4}$
$\lambda_{\min} = \frac{4}{R_{H}}$
Given $R_{H} = 1\cdot097 \times 10^{7}\text{ m}^{-1}$:
$\lambda_{\min} = \frac{4}{1\cdot097 \times 10^{7}\text{ m}^{-1}} \approx 3\cdot646 \times 10^{-7}\text{ m} = 364\cdot6\text{ nm}$
Final answer: 3.65 \times 10^{-7} m
Final answer: 3.65 \times 10^{-7} m
From ISC 2026 Physics Paper 1, question 19(i).