Correct Answer - A
For ionisation `n_(2) = prop ` Hence
`Delta E = R((1)/(n_(1)^(2)))`
b . The ionisation energy in proportion to the square of the charge in the nucleuws as is evident from the expression
`R = (2pi^ (2)m((Ze^(2))/(4 pi epsilon_(0)))^(2))/(h^(3)c)`
`Delta barE _(1) = R`
c. The correct expression is `Delta p Delta s ge (h)/(4pi)`
d. For a multi -electron atom the energy of an orbital depends on both principle and azimuthal quantum number .The largwe tyhe value of `n = l`, the large the value of n, the large the energy