Answer:
The correct option is 'c':electron,proton,helium nucleus
Explanation:
The De-Broglie's wavelength of particle is given by

Thus we can see that wavelength is inversely related to mass of the particle since 'h' (Plank's constant) and velocity is same for all the particles
Thus we conclude that the the lightest particle will have the most wavelength
Electron being the lightest of the 3 particles will have the largest wavelength thus the correct option is 'c'. Since electron has the largest wavelength followed by proton and the least wavelength among the 3 is of helium.
Answer:
λ = a
Explanation:
This is a diffraction exercise that is described by the expression
a sin θ = m λ
sin θ = m λ/ a
the first zero of the diffraction occurs for m = 1
sin θ = λ / a
angles are generally very small and are measured in radians
sin θ = θ = y / x
we substitute
the width of the central maximum is twice the distance to zero
w = 2y
in the exercise indicate that this width is equal to twice the distance to the screen (2x)
W = 2x
2y = 2x
we substitute
1 = λ/ a
λ = a
we see that the width of the slit is equal to the wavelength used.