Each of the following questions contains statement given in two columns, which have to be matched. The answers to these questions have to be appropriately dubbed. If the correct matches are $\mathrm{A}-\mathrm{q}, \mathrm{s}, \mathrm{B}-\mathrm{p}, \mathrm{r}, \mathrm{C}-\mathrm{q}, \mathrm{s}$ and \(\mathrm{D}-\mathrm{s}\) then the correctly dabbled matrix will look like the one shown here: Match the statements of column I with that of column II. Column-I Column-II (A) \(\propto\) -particle and proton have same \(\mathrm{K} . \mathrm{E}\). \((\mathrm{p}) \lambda_{\mathrm{p}}=\lambda_{\propto}\) (B) \(\propto\) -particle has one quarter \(\mathrm{K} . \mathrm{E}<\) than that (q) \(\lambda_{\mathrm{p}}>\lambda_{\propto}\) (C) proton has one quarter \(\mathrm{K} . \mathrm{E}\). than that (r) \(p_{p}=p_{\propto}\) (D) \(\propto\) -particle and proton same velocity (s) \(\mathrm{p}_{\propto}>\mathrm{p}_{\mathrm{p}}\)

Step by step solution

01

Match A from Column I

For Column I - (A) \(\alpha\)-particle and proton have the same kinetic energy (K.E). Using the de Broglie wavelength formula: \(\lambda = \frac{h}{p}\), where \(h\) is the Planck's constant and \(p\) is the momentum. For the same kinetic energy, the heavier particle has greater momentum. Recalling that an \(\alpha\)-particle is made up of 2 protons and 2 neutrons, we can see that the \(\alpha\)-particle is heavier than a proton. Thus, the best match for statement A would be: Column II - (s) \(p_{\alpha} > p_p\) (as the \(\alpha\)-particle has greater momentum than the proton).

02

Match B from Column I

For Column I - (B) \(\alpha\)-particle has one quarter K.E than that. For a fixed wavelength (\(\lambda\)), the momentum \(p = \frac{h}{\lambda}\) and kinetic energy \(K.E. = \frac{p^2}{2m}\) are inversely proportional to the mass of the particle. As the kinetic energy of an \(\alpha\)-particle is one quarter that of the proton, it has a longer wavelength. Thus, the best match for statement B would be: Column II - (q) \(\lambda_p > \lambda_\alpha\)

03

Match C from Column I

For Column I - (C) proton has one quarter K.E than that. Following the same analysis as in Step 2, if the proton has one quarter K.E., its wavelength will be longer than that of the \(\alpha\)-particle. Thus, the best match for statement C would be: Column II - (p) \(\lambda_p = \lambda_\alpha\) (since the wavelength is the same for both particles).

04

Match D from Column I

For Column I - (D) \(\alpha\)-particle and proton have the same velocity. Recalling that the momentum is given by \(p = mv\) where \(m\) is the mass of the particle and \(v\) is its velocity, we can see that if both particles have the same velocity, the heavier particle, which is the \(\alpha\)-particle, will have greater momentum. Thus, the best match for statement D would be: Column II - (r) \(p_p = p_\alpha\) (since both particles have the same momentum). After all steps, the correctly dabbed matrix looks like this: (A) \(\propto\) -particle and proton have same K.E. => (s) (B) \(\propto\) -particle has one quarter K.E<$ than that => (q) (C) proton has one quarter K.E. than that => (p) (D) \(\propto\) -particle and proton same velocity => (r)

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