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Answer the following.

Draw a labelled graph showing the variation of binding energy per nucleon with the mass number (A)…

Physics20265 marksGraph
(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

(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)

AI
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)

AI
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

Nuclei

From ISC 2026 Physics Paper 1, question 19(i).