An $\alpha$- particle of $5\ MeV$ energy strikes with a nucleus of uranium at stationary at an scattering angle of $180^o$. The nearest distance upto which $\alpha$- particle reaches the nucleus will be of the order of

Hydrogen atom is excited from ground state to another state with principal quantum number equal to $4$. Then the number of spectral lines in the emission spectra will be

Ratio of longest wave lengths corresponding to Lyman and Balmer series in hydrogen  spectrum is

In an alpha particle scattering experiment distance of closest approach for the $\alpha$ particle is $4.5 \times 10^{-14} \mathrm{~m}$. If target nucleus has atomic number $80$ , then maximum velocity of $\alpha$-particle is . . . . .. $\times 10^5$ $\mathrm{m} / \mathrm{s}$ approximately.

$\left(\frac{1}{4 \pi \epsilon_0}=9 \times 10^9 \mathrm{SI}\right.$ unit, mass of $\alpha$ particle $=$ $\left.6.72 \times 10^{-27} \mathrm{~kg}\right)$

Give the relationship between impact parameter and scattering angle. 

In the spectrum of hydrogen atom, the ratio of the longest wavelength in Lyman series to the longest wavelength in the Balmer series is