The frequency of radiation emitted when the electron falls from n = 4 to n = 1 in a hydrogen atom will be (given ionization energy of $H=2.18\times {10}^{-18}ato{m}^{-1}$ and $h=6.606\times {10}^{-34}j-s$)

Q. 13. The frequency of radiation emitted when the electrons falls from n=4 to n=1 in a hydrogen atom will be[given ionisation energy of H=2.18 into 10 power minus 18 j/atom and h= 6.625into 10 minus 34

Q. The frequency of radiation emitted when the electron falls from $n=4$ to $n=1$ in a hydrogen atom will be (Given : ionisation energy of $H=2.18\times {10}^{-18}J{atom}^{-1}$ and $h=6.626\times {10}^{-34}J-s$)

Q. The frequency of radiation emitted when the electron falls from $n=4$ to $n=$ lina hydrogen atom will be (Given ionisation energy of $H=2.18{10}^{-18}Jato{m}^{-1}$ and $h=6.625\times {10}^{-34}Js)$

Q. Q.] Determine the frequency[f] of revolution of an electron in the second orbit of helium atom.
ANS:- Data :
z = 2, n= 2, n^th =2^nd , v = 2.18 x 10⁶ ms^-1 , find frequency(f).
Now,
n= v/ 2 pie r
n = 2.18 x 10⁶ ms^-1 / 2 x 3.14 x 0. 529 x 10^-10 m
n = [ 2.18 x 10⁶ / 2 x 3.14 x 0.529 ] s^-1 .
n = [ 2.18 x 10¹⁶ / 2 x 3.14 x 0.529 ] s^-1
n = 6.562 x 10¹⁵ [ f of atom in ground state of hydrogen atom ]
Hence,
n = 6.562 x 10¹⁵ x 2
n = 13.124 x 10⁵ (Hz) .
[ is this correct method ? am i in right direction? ]

Q. The threshold frequency of a metal is 7.0 × 1015 s-1. Calculate the kinetic energy of an electron emitted when radiation of frequency 1×10¹5 s-1 hits the metal