Question

The ionization energy of hydrogen atom is 13.6 eV. Hydrogen atoms in the ground state are excited by electromagnetic radiation of energy 12.1 eV. How many spectral lines will be emitter by the hydrogen atom?

Solution

Correct option is

Two

When an electron in the ground state receives 12.1 eV of energy, it jumps to a level where its energy = 13.6 – 12.1 = 1.5 eV. This corresponds to the third excited state corresponding to n = 3. It can have two transitions, namely from n = 3 to n = 2 and from n = 3 to n = 1. Hence two spectral lines will be emitted.

SIMILAR QUESTIONS

Q1

Which of the following transition in a hydrogen atom emits the photon of lowest frequency?  

Q2

How many times does the electron go around the first Bohr orbit in one second? 

Q3

The electron orbits with principal quantum numbers n > 3 were not allowed, the number of possible elements would be  

Q4

According to Bohr’s theory, the energy of an electron in the nth orbit of an atom of atomic number Z is proportional to 

Q5

The dimensions of the Rydberg’s constant are

Q6

In terms of the Rydberg’s constant R, the minimum wavelength in the Lyman series is  

Q7

The wavelength of the second line of Balmer series is 486.4 nm. What is the wavelength of the first line of Lyman series?

Q8

The wavelength of the first line in Balmer series in the hydrogen spectrumis . What is the wavelength of the second line. 

Q9

The energy needed to ionize the hydrogen atom in the first excited state is 3.4 eV. What is the ionization potential of hydrogen in the ground state?

Q10

If an orbital electron of the hydrogen atom jumps from the ground state to a higher energy state, its orbital speed reduces to half its initial value. If the radius of the electron orbit in the ground state is r, then the radius of the new orbit would be