## Question

### Solution

Correct option is

13.6 V

Let E be the ionization energy of the hydrogen atom. Now for the first excited state n = 2.

Therefore,

= 13.6 eV.

#### SIMILAR QUESTIONS

Q1

The shortest wavelength in the :Lyman Series is 911.6 Å. Then the longest wavelength in the Lyman’s Series is

Q2

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

Q3

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

Q4

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

Q5

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

Q6

The dimensions of the Rydberg’s constant are

Q7

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

Q8

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

Q9

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

Q10

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?