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A single electron orbikt around a statio...

A single electron orbikt around a stationary nucleus of charge `+ Ze` when Z is a constant and e is the magnitube of the electronic charge if `47.2 eV excite the electron from the second bohr orbit to the third bohhr orbit final
(i) The volue of Z
(ii) Tyhe energy requiredto nucleus the electron from the third to the fourth bohr orbit
(iii) The wavelength of the electronmagnetic radiation required to remove the electron from the first bohr orbit to inlinity
(iv) The energy potential energy potential energy and the angular momentum of the electron in the first bohr orbit
(v) The radius of the first bohr orbit (The ionization energy of hydrogen atom ` = 13.6 eV ` bohr radius `= 5.3 xx 10^(-11) matre` velocity of light `= 3 xx 10^(8) m//sec` planks 's constant ` = 6.6 xx 10^(-34)` jules - sec )

Text Solution

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The correct Answer is:
A, C, D

`E_(2) = (13.6)/(4) Z^(2) , E_(2) =(13.6)/(9) Z^(2)`
E_(3) - E_(2) = -13.6 Z^(2)((1)/(9) - (1)/(4)) = + (13.6 xx 5)/(36) Z^(2)`
But ` E_(3) - E_(2) =47.2 eV` (Given)
`:. (13.6 xx 5)/(36) Z^(2)`= 47.2 ,. Z = (sqrt(47.2 xx 36))/(13.6 xx 5 ) = 5 `
(ii)` E_(0) = (=(13.6)/(16) Z^(2)`
`:. E_(4) - E_(3) = - 13.6Z^(2) [(1)/(16) - (1)/(9)] = -13.6Z^(2) [(9 -16)/(9 xx 16)]`
= (+ 13.6 xx 25 xx 7 )/(9 xx 16) = 16.53 eV`
(iii)`E_(1) = (13.6)/(1) xx 25 = - 340eV`
`E= E_(oo) - E_(1) = 340eV = 340 xx 16 xx 10^(-19) J (E_(oo)eV]`
But E = (hc)/(lambda)`
`:. lambda = (hc)/(E)= (6.6 xx10^(-34) xx 3 xx10^(8))/(340 xx 10^(-19) xx 1.6 ) = 3.65 xx 10^(-19) m `
Total energy of ist orbit `= - 340 eV`
we know that `- (T.E) = K.E. `[in case of revolving around nuclues ]
and `2T.E = pE. `
:. K.E. = 340 eV , PE = - 680 eV`
KEY CONCEPT :
Aungle momentum in 1st orbuit :
Account to Bohr 's postualate ,
` mor = (nh)/(2 pi) `
`for n = 1 , `
mor = (h)/(2 pi) = (6.6 xx 10^(-34))/(2 pi) = 1.05 xx 10^(-34) J- s `
(v) Radius of first Bohr orbit
` r_(1) = (5.3 xx 10^(-11))/(Z) = (5.3 xx 10^(-11))/(5) `
`= 1.06 xx 10^(-11)m `
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