Question

Electrons are emitted from an electron gun at almost zero velocity and are accelerated by an electric field E through a distance of 1.0 m. The electrons are now scattered by an atomic hydrogen sample in ground state. What should be the minimum value of E so that red light of wavelength 656.3 nm may be emitted by the hydrogen?

Answer 1

The minimum of electric field (E)  so that red light of wavelength 656.3 n-m may be emitted by the hydrogen is 12.1 V per meter

Explanation:

It is shown that the distance moved by the electron, d = 1.0 m

Red light’s wavelength, $\lambda = 656.3 \times 10^-9 m$

In Balmer series, the given wavelength lies, the changeover that needs minimum energy is to $n_2$ = 2 from $n_1$ = 3.

This transition’s energy will be equal to the energy $E_1$ that will be needed for the ground state transition to n = 3.

$E_1 = 13.6(\frac{1}{n_1^2} - \frac{1}{n_2^2} )$

$E_1 = 13.6 (\frac{1}{1} - \frac{1}{9} )$

$E_1$ =  12.09 eV

Therefore, V = 12.09 V

Electric field, E = V × d

E = 12.09 ×1.0

E = 12.09 V/m

Therefore, the electric field’s minimum value = 12.1 V/m  so that red light of wavelength 656.3 n-m may be emitted by the hydrogen

Answer 2

Answer:

Above answer is correct ..............