The distance of closest approach when an alpha particle of kinetic energy $$6.5 \mathrm{~MeV}$$ strikes a nucleus of atomic number 50 is
If an electron in a hydrogen atom jumps from the third orbit to the second orbit, it emits a photon of wavelength $$\lambda$$. When it jumps from the second to the first orbit, the corresponding wavelength of the photon will be
An electron has a mass of $$9.1 \times 10^{-31} \mathrm{~kg}$$. It revolves round the nucleus in a circular orbit of radius $$0.529 \times 10^{-10} \mathrm{~m}$$ at a speed of $$2.2 \times 10^6 \mathrm{~ms}^{-1}$$. The magnitude of its angular momentum is
$$ \text { If the nuclear radius of }{ }^{27} \mathrm{Al} \text { is } 3.6 \text { fermi, the nuclear radius of }{ }^{125} \mathrm{Fe} \text { is } $$