Atomic StructureHard

Question

A hydrogen like atom of atomic number Z is in an excited state of quantum number 2n. It can emit a maximum energy photon of 204 eV. If it makes a transition to quantum state n, a photon of energy 40.8 eV is emitted. The value of n will be

Options

A.1
B.2
C.3
D.4

Solution

Let ground state energy (in eV) be E1.
Then from the given condition
E2n -  E1 = 204eV or - E1 = 204eV
⇒ E1 = 204 eV   .....(i)
and E2n - En = 40.8 eV
⇒  = 40.8 eV
From equation (i) and (ii),
= 5 ⇒  n = 2

Create a free account to view solution

View Solution Free
Topic: Atomic Structure·Practice all Atomic Structure questions

More Atomic Structure Questions

The transition from the state n = 3 to n = 1 in a hydrogen like atom results in ultraviolet radiation.Infrared radiation...The threshold wavelength for photoelectric emission from a material is $5200\ \text{Å}$. Photoelectrons will be emitted ...If the electron in hydrogen orbit jumps from third orbit to second, the wavelength of the emitted radiation is given by ...A solid which is transparent to visible light and whose conductivity increases with temperature is formed by...Energy required for the electron excitation in Li++ from the first to the third Bohr orbit is :...