Chapter 19: Problem 167
The binding energy per nucleon for the parent nucleus is \(E_{1}\) and that for the daughter nuclei is \(E_{2}\). Then (A) \(E_{1}=2 E_{2}\) (B) \(E_{2}=2 E_{1}\) (C) \(E_{1}>E_{2}\) (D) \(E_{2}>E_{1}\)
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Chapter 19: Problem 167
The binding energy per nucleon for the parent nucleus is \(E_{1}\) and that for the daughter nuclei is \(E_{2}\). Then (A) \(E_{1}=2 E_{2}\) (B) \(E_{2}=2 E_{1}\) (C) \(E_{1}>E_{2}\) (D) \(E_{2}>E_{1}\)
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If \(13.6 \mathrm{eV}\) energy is required to ionize the hydrogen atom, then the energy required to remove an electron from \(n=2\) is \(\quad\) [2002] (A) \(10.2 \mathrm{eV}\) (B) \(0 \mathrm{eV}\) (C) \(3.4 \mathrm{eV}\) (D) \(6.8 \mathrm{eV}\)
If the binding energy per nucleon in \({ }_{3}^{7} \mathrm{Li}\) and \({ }_{2}^{4} \mathrm{He}\) nuclei are \(5.60 \mathrm{MeV}\) and \(7.06 \mathrm{MeV}\) respectively, then in the reaction \(p+{ }_{3}^{7} \mathrm{Li} \rightarrow 2{ }_{2}^{4} \mathrm{He}\) energy of proton must be (A) \(28.24 \mathrm{MeV}\) (B) \(17.28 \mathrm{MeV}\) (C) \(1.46 \mathrm{MeV}\) (D) \(39.2 \mathrm{MeV}\)
When photons of energy \(5 \mathrm{eV}\) strike the surface of a metal \(A\), the ejected photoelectrons have maximum kinetic energy \(K_{A} \mathrm{eV}\) and de Broglie wavelength \(\lambda_{A}\). The maximum kinetic energy of photoelectrons liberated from another metal \(B\) by photons of energy \(5.30 \mathrm{eV}\) is \(K_{B}=\left(K_{A}-1.5\right) . \mathrm{eV}\). If the de Broglie wavelength of these photoelectrons is \(\lambda_{B}=2 \lambda_{A}\), then find \(K_{A}\) and \(K_{B}\).
The work function \(W_{A}\) for photoelectric material \(A\) is \(2 \mathrm{eV}\) and \(W_{B}\) for another photoelectric material \(B\) is \(4 \mathrm{eV}\). If photons of energy \(E_{A}\) strike the surface of \(A\), the ejected photoelectrons have a minimum de Broglie wavelength and photons of energy \(E_{B}\) strike the surface \(B\), the ejected photoelectrons also have a minimum de Broglie's wavelength. If \(E_{B}-E_{A}=\) \(0.5 \mathrm{eV}\) and \(V_{A}\) and \(V_{B}\) are the respective stopping potentials, find \(V_{A}-V_{B}\).
Energy required for the electron excitation in \(\mathrm{Li}^{++}\) from the first to the third Bohr orbit is (A) \(12.1 \mathrm{eV}\) (B) \(36.3 \mathrm{eV}\) (C) \(108.8 \mathrm{eV}\) (D) \(122.4 \mathrm{eV}\)
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