Question

Photon having energy equivalent to the binding energy of 4th state of He+ atom is used to eject an electron from the metal surface of work function 1.4 eV. If electrons are further accelerated through the potential difference of 4V then the minimum value of De–broglie wavelength associated with the electron is :

• A. 1.1 Å
• B. 5 Å
• C. 9.15 Å
• D. 11 Å

Answer 1

Answer: (B)

5 Å

Answer 2

Total energ$= \frac{13.6Z^{2}}{n^{2}} = \frac{13.6(Z)^{2}}{(4)^{2}} = 3.4eV$

Now K.E. = 3.4 – 1.4 = 2 eV

Now, Total energy = 2 + 4 = 6 eV

i.e. potential = 6 V

For electron