An electron in the first excited state of if atom obserbed a photon and further excited .The de broglie wavelength of the electron in this state is found to be `13.4Å` Find the wavelength of the photon abserbed by the electron in angstroms Also find the longest and the shorted wavelength emitted when this electron de-excited back to the ground state

Explanation

The electron from `n = 2 ` absorted a photon and is further excited to a higher energy lowest (say n)
The electron in this frist level (n) has a de broglie wavelength `(lambda) = 13.4 Å`

`lambda_(e) = (h)/(m_(e)v_(e))`
`and v_(e) = 2.18 xx 10^(6)(Z)/(n) m s^(-1)`
`[v_(A)` "is the velocity of the nth orbit"]
`rArr (6.626 xx 10^(-34))/((13.4 xx 10^(-10)) xx (9.1 xx 10^(-31))) = 2.18 xx 10^(6) xx (1)/(n)`
`rArr n = 4`
Now find the wavelength of the photon respensible for the excitation from `= mn = 2 to n = 4`
Using the relation
` overset(Delta E)underset((2 rarr 4)) = 2.18 xx 10^(-18) Z^(2)((1)/(n_(1)^(2)) - (1)/(n_(2)^(2)))`
`= 4.09 xx 10^(-19)J [n_(1) = 2, n_(2) = 4, Z = 1]`
`rArr = (hc)/(lambda) = 4.09 xx 10^(-19)`
`lambda = 4963.1 Å`
The longest wavelength emitted when this electron (from `n = 4`) falls back to the ground state with correspond to the minimum energy transition
THe transition corresponding to the minimum energy wioll be `4 rarr 3`
The transition corresponding to the maximum energy will be `4 rarr 1`
`Delta E = E_(photon) = (hc)/(lambda) = hv`
`rArr Delta E = (1)/(lambda_("photon")) or Delta E = V_("photon")`
Using the same relation
`Delta E _((4 rarr 3)) = 2.18 xx 10^(-18) Z^(2) ((1)/(n_(1)^(2)) - (1)/(n_(2)^(2))) [ n_(1) = 3, n_(2) = 4, Z = 1]`
`rArr Delta E _((4 rarr 3))= 1.06 xx 10^(-19) J`
`Delta E = E_("photon") = (hc)/(lambda) = 1.06 xx 10^(-19) J`
`rArr = 18752.8 Å`
Shortest wavelength ` 4rarr 1`
`Delta E _((4 rarr 3))= 2.18 xx 10^(-19) xx 1^(2) xx ((1)/(1^(2)) - (1)/(4^(2)))`
`= 2.04 xx 10^(-19) J`
`rArr Delta E _((4 rarr 1))=E_("photon") = (hv)/(lambda)`
`rArr lambda = 973.2 Å`