A hydrogen like atom of atomic number Z is in and excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, photon of energy 40.8 eV is emitted. Find n , Z and the gound state energy (in eV) for this atom, Also calculate the minimum energy (in eV) that can be emitted by this atom during de-exitation, Ground state energy of hydrogen atom is `13. 6 eV`

Options

A. `10.32 Mhz`
B. `10.61 kHz`
C. `5.31 MHz`
D. 5.31 kHz

Answer

B

Explanation

`therefore tau=RC = 100 xx 10^(3) xx 250 xx 10^(-12)`s
`=2.5 xx 10^(7) xx 10^(-12) s=2.5 xx 10^(-6)`s
The higher frequency which can be detected with tolerable distortion is
`f=1/(2pim_(a)RC)= 1/(2pixx0.6xx2.5xx10^(-6))`Hz
`=(100 xx 10^(4))/(25 xx 12pi)`Hz
`=(4/12pi xx 10^(4)` Hz = 10.61 kHz