Question

A hydrogen like species ( atomic number z) is present in a higher excited state of quantum number n .This excited atom can make a transition to the first excited state by successive emission of two photons of energy 10.20 eV and 17.0 eV respectively .Alternatively, the atom from the same excited state can make transition to the second excited state by successive emission of two photons of energy 4.25 eV and 5.95 eV respectively. Determine the value of Z.

Answer 1

Here we know that

One photon emits 10.20 eV and other photon emits 17 eV

Hence the total energy emitted is = 10.20 + 17 =27.20 eV

There fore this is the energy emitted by the photon from transition from n=n to n=2

We know that energy released = hc/lambda ______(1)

Where h is planck’s conctant and c is velocity of light

Lamba is wavelength

Therefore 27.20=hc/lambda

We know rydbergs equation

Which is given as –

1/lambda= Rh[1/n1^2 – 1/n2^2]*Z^2

Where Rh is rydbergs constant

We can write above equation as (multiply by hc on both sides)

Hc/lambda = (hc)Rh*[1/2^2-1/n^2]*Z^2___________(2)

i.e. 27.20=(hc)Rh[1/4-1/n^2]*Z^2

let us mark this equation as (3)

and the total energy liberated during the transition of electrons from n=n to n=3 is =(4.25+5.95)=10.20eV

from (1)

we get

hc/lambda=10.20 eV

from (2) we get

10.20=(hc)Rh [1/3^2-1/n^2]*Z^2

10.20==(hc)Rh [1/9-1/n^2]*Z^2

Let us mark this as equation (4)

Dividing equation (3) by (4) we get

27.20/10.20=[1/4-1/n^2]/  [1/9-1/n^2]

2.67==[1/4-1/n^2]/  [1/9-1/n^2]

[1/9-1/n^2]2.67=[1/4-1/n^2]

2.67/9-2.67/n^2=1/4-1/n^2

-2.67/n^2+1/n^2=1/4-2.67/9

-2.67/n^2+1/n^2=0.25-0.2967

-2.67/n^2+1/n^2=-0.0467

-1.67/n^2=-0.0467

1.67/n^2=0.0467

n^2=1.67/0.0467

n^2=35.76 which is approximately equal to 36

therefore n=6

put the value of n in equation (3) and (4)

and then we get Z=3 for solving equation 3and 4 u would need the value of planck’s constant which is equal to 6.626176 x 10-34 joule-seconds. And we the c is velocity of light

and thus on solving both equations we get z=3

Hope this helped u :-)

Answer 2

We know that energy of the electron in the nth energy state in an atom is given by Bohr Sommer-field model as below. This include Kinetic and electrostatic energies.   Let Z = atomic number.

E = - 13.6 Z² / n²   eV.

ground state means : n =1.  First excited state means 2.  The second excited state means n =3.

Let the intermediate energy state in the first case be x.  Adding the energies of the two photons emitted, gives energy gap between states n and 2. So

$\frac{Z^2}{4}-\frac{Z^2}{n^2}=\frac{10.20+17.0}{13.6}=2,\ .....(1)$

Let the intermediate energy state in the 2nd case be y.  Adding the energies of the two photons emitted is energy gap between state n and 3 :

$\frac{Z^2}{9}-\frac{Z^2}{n^2}=\frac{4.25+5.95}{13.6}=\frac{3}{4},\ ...(2)$

Subtracting (1) from (2), we get :

$\frac{5}{36}Z^2=\frac{5}{4}\\\\Z^2=9,\ \ Z=3$

Hence the atomic number is 3.  Is that Boron?

Now can also find the intermediate energy states x and y.