## Question

### Solution

Correct option is Energy state –3.4 eV corresponds to a level n given by which gives n = 2.

Now, orbital speed .

Hence the orbital speed in the excited state is .

#### SIMILAR QUESTIONS

Q1

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

Q2

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

Q3

The dimensions of the Rydberg’s constant are

Q4

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

Q5

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

Q6

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

Q7

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?

Q8

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?

Q9

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

Q10

The ratio of the wavelengths of the longest wavelength lines in the Lyman and Balmer series of hydrogen spectrum is