Question

In young's double slit experiment, a coherent parallel stream of electrons, accelerated by a potential difference V = 45.5 volt is used to obtain interference pattern. If slits are separated by a distance d = 66.3 mm and distance of screen is D = 109.2 cm from plane of slits, calculate distance between two consecutive maxima on the screen : [mass of electron = 9.1 × 10^–31 kg Plank constant h = 6.63 × 10^–34 J-s]

• A. 6 mm
• B. 2 mm
• C. 9 mm
• D. 3 mm

Answer 1

Answer: (D)

3 mm

Answer 2

$\frac{1}{2}$mv² = eV

v = $\sqrt{\frac{2eV}{m}}$

l =$\frac{h}{mv}$=$\frac{h}{\sqrt{meV}}$

b = $\frac{\lambda D}{d}$=$\frac{hD}{d\sqrt{2meV}}$= 3 × 10^–6 m = 3 mm