Question

A hydrogen-like atom (atomic number Z) is in a higher state of Quantum number n. This excited atom can make a transition to the first state by successfully emitting two photons of energies ${ 10.20 eV }$ and ${ 17.00 eV }$. Alternatively, the atom from the same excited state can make a transition to the energy ${ 4.24 eV }$ and ${ 5.95 eV }$. If the excited level of the second transition is ${ 3 }$. The value of the atomic number (Z) is _______.
(A) ${ 2 }$
(B) ${ 4 }$
(C) ${ 6 }$
(D) ${ 3 }$

Answer 1

Hint: It is the number of protons in an atom. It is equal to the number of electrons as well. The Atomic number = Number of electrons. The atomic number is found in the nucleus as the protons are present in the nucleus.

Complete step by step solution:
It is given that,
n = ${ 2 }$
For the transition from the excited state,n to the first excited state, ${ n }_{ 1 }{ =2 }$
${ E }_{ n }{ -E }_{ 2 }$ = ${ 10+17 = 27eV }$……….(1)
For the transition from the excited state,n to the second excited state, ${ n }_{ 2 }{ =3 }$
${ E }_{ n }{ -E }_{ 3 }$ = ${ 4.25+5.95 = 10.2eV }$……….(2)
From equation (1) and (2), we get
${ E }_{ 3 }{ -E }_{ 2 }$ = ${ 17 }$........(3)
As we know,
${ \Delta E=R }_{ h }{ \times { z }^{ 2 } }\times { hc[1\div (n }_{ 1 }{ ) }^{ 2 }{ -1\div ({ n }_{ 2 } }{ ) }^{ 2 }{ ] }$
where ${ R }_{ h }$ = Rydberg’s constant
z= atomic number
h= Planck’s constant
c= velocity of light
${ \Delta E } { =13.6Z }^{ 2 }{ [1\div (n)_{ 1 } }^{ 2 }{ -[1\div (n) }_{ 2 }^{ 2 }{ ] }$
${ \Delta E }_{ 2 }{ =13.6Z }^{ 2 }{ [1\div (2) }^{ 2 }{ -[1\div (n) }^{ 2 }{ ] }$...........(4)
${ \Delta E }_{ 3 }{ =13.6Z }^{ 2 }{ [1\div (3) }^{ 2 }{ -[1\div (n) }^{ 2 }{ ] }$............(5)
Putting the values in equation (3), we get
${ E }_{ 3 }{ -E }_{ 2 }{ =-13.6{ Z }^{ 2 } }{ \div 9+13.6Z^{ 2 }\div 4=17 }$
${ Z }^{ 2 }{ =9 }$
${ Z=3 }$
Now, put the value of Z in equation (1), we get
${ -13.6\times { Z }^{ 2 }\div { n }^{ 2 } }{ +13.6\times { Z }^{ 2 }\div 4 }{ =27 }$
${ -13.6\times 9\div { n }^{ 2 } }{ +13.6\times 9\div 4 }{ =27 }$
${ 1\div n }^{ 2 }{ =0.028 }$
${ n }^{ 2 }{ =35.7 }$
n = ${ 6 }$
Hence, the value of the atomic number (Z) is ${ 3 }$.

The correct option is D.

Note: The possibility to make a mistake is that ${ n }_{ 1 }$ and ${ n }_{ 2 }$ are the transitions from ground to first and second excited state respectively and you have to subtract the ground state energy from excited-state energy.