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perin

an electron in a hydrogen LIKE species is associated with a wave length of 6.67 A⁰. wave no. of lyman's series of that specie is 438708 cm^-1. the species is excited electrically to some higher quantum state which has 3.808 eV more than the 2nd excited state. R = 109677 cm^-1

1 what is Kinetic energy of electron? 3.38 eV
2 what is PE of e- ? -6.78 eV
3 what will b wavelength of photon emitted during the shortest wavelength of transition from the mentioned excited state? 237.4 A⁰

perin

sumbdy answer pleeeeez.

sahilgupta_iit

is the wave no. given is that of the photon used?

kshitijgunit

remember :--
K.E=-E
and
P.E=2E, where E is the difference in energy levels...
id love to derieve it for you but it requires variables tht my keyboard doesnt posses...if ur an ISC student, theres a similar question on Pg . 996 in the book by kumar mittal..

kshitijgunit

ohkk here it is anyways,
K.E=(1/2)MVsquare...
now MVsq=Z*esq/4*pi*epsylon*r (from equating the coloumbic force of attraction to the centripetal force)
therefore, K=Z*esquare/8*pi*epsylon*r
PE= Z*esquare/4*pi*epsylon*r (from the electrostatic potential energy concept)
now E= KE + PE, therefore E=/4*pi*e0*(esquare/2r-qsquare/r)
=- esq*Z/8pie0r = - K, and = 2times P.E

kshitijgunit

hope it makes sense to u lol

sahilgupta_iit

u haven't answered yet Perin

sahilgupta_iit

i asked u whether the wave no. given is that of photon used.

sachin_gupta1991

Perin, I have answered ur question completely when u had asked it earlier.

U can look there to get the solution.

sachin_gupta1991

1) To find the kinetic energy, use the formula

=h/mv and then calculate kinetic energy by 1/2mv².

= 6.67 x 10^-10 = 6.63 x 10^-34

----------------------------------------

9.1 x 10^-31 x v

which on simplification gives v=0.109 x 10⁷ m/s.

K.E. = 1/2mv² where m = mass of electron

On putting the values and getting the answer in joule as

5.4 x 10^-19 J , convert it into eV to get 3.38 eV.

2) Now,

as K.E. = kZe²

------

2r

and P.E.= - kZe²/ r

we have P.E.= - 2 K.E.

Thus, P.E.= -6.78 eV.

3) Wavenumber

-------------------------- = Z² {1- 1/

²}

Rydberg's constant

From this u will get Z=2.

Now, E= -13.6 Z²/N² EV

Using this or rydberg's equation, wavelength of photon emitted during the shortest wavelength of transition from the excited state can be found out easily.

perin

sorry sahil but i myself didn't get wat's happening in this question...

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