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Chapter 18: Problem 2464
In Bohr model the hydrogen atom, the lowest orbit corresponds to (A) Infinite energy (B) zero energy (C) The minimum energy (D) The maximum energy
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An \(\alpha\) -particle of energy \(5 \mathrm{MeV}\) is scattered though \(180^{\circ}\) by a fixed uranium nucleus. The distance of the closet approach is of the order of (A) \(10^{-8} \mathrm{~cm}\) (B) \(10^{-12} \mathrm{~cm}\) (C) \(10^{-10} \mathrm{~cm}\) (D) \(10^{-15} \mathrm{~cm}\)
Complete the reaction ${ }_{0} \mathrm{n}^{1}+{ }_{92} \mathrm{U}^{235} \rightarrow{ }_{56} \mathrm{Ba}^{144}+{ }_{\mathrm{Z}} \mathrm{X}^{\mathrm{A}}+3\left({ }_{0} \mathrm{n}^{1}\right)$ (A) \(_{36} \mathrm{Kr}^{90}\) (B) \(_{36} \mathrm{Kr}^{89}\) (C) \(_{36} \mathrm{Kr}^{91}\) (D) \(_{36} \mathrm{Kr}^{92}\)
In each of the following question match column -I and column -II. Select correct Answer. (a) Bohr atom model (p) fixed for the atom (b) Ionization potential (q) Nucleus (c) Rutherford atom model (r) stationary orbits (d) Thomson atom model (s) In atom positive and Negative charge are distributed uniformly (A) $\mathrm{a} \rightarrow \mathrm{s}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{p}$ (B) $\mathrm{a} \rightarrow \mathrm{r}, \mathrm{b} \rightarrow \mathrm{p}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{d} \rightarrow \mathrm{s}$ (C) $\mathrm{a} \rightarrow \mathrm{p}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{c} \rightarrow \mathrm{s}, \mathrm{d} \rightarrow \mathrm{q}$ (D) $\mathrm{b} \rightarrow \mathrm{p}, \mathrm{c} \rightarrow \mathrm{q}, \mathrm{b} \rightarrow \mathrm{r}, \mathrm{d} \rightarrow \mathrm{s}$
If the binding energy Per nucleon in \({ }_{3}^{7} \mathrm{Li}\) and ${ }^{4}{ }_{2}\( He nuclear is \)5.6 \mathrm{MeV}\( and \)7.06 \mathrm{MeV}$ respectively, then in the reaction $\mathrm{P}+{ }_{3} \mathrm{Li} \rightarrow 2\left({ }_{2}^{4} \mathrm{He}\right)$ (P here retrent Proton) energy of Proton must be (A) \(1.46 \mathrm{MeV}\) (B) \(39.2 \mathrm{MeV}\) (C) \(17.28 \mathrm{MeV}\) (D) \(28.24 \mathrm{MeV}\)
The masses of neutron and Proton are \(1.0087\) amu and \(1.0073\) amu respectively. It the neuron and Protons combins to form Helium nucleus of mass \(4.0015\) amu then binding energy of the helium nucleus will be (A) \(14.2 \mathrm{MeV}\) (B) \(28.4 \mathrm{MeV}\) (C) \(27.3 \mathrm{MeV}\) (D) \(20.8 \mathrm{MeV}\)
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