Answer:
The light used has a wavelenght of 4.51×10^-7 m.
Explanation:
let:
n be the order fringe
Ф be the angle that the light makes
d is the slit spacing of the grating
λ be the wavelength of the light
then, by Bragg's law:
n×λ = d×sin(Ф)
λ = d×sin(Ф)/n
λ = (3.2×10^-4 cm)×sin(25.0°)/3
= 4.51×10^-5 cm
≈ 4.51×10^-7 m
Therefore, the light used has a wavelenght of 4.51×10^-7 m.
There is no sound in space because there is no medium to move it. However i think the big bang can still been seen in the form of background radiation but i am unsure.
Answer: 2.068*
m
Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.
workdone= qV
energy = hc/λ
q=magnitude of an electronic charge= 1.602*
h= planck constant = 6.626*
c= speed of light =2.998* 
v= potential difference= 6*
λ= wavelength=unknown
by making λ subject of formulae we have that
λ= 
λ = 6.626*
* 2.998*
/ 1.602*
* 6*
λ = 
by doing the necessary calculations, we have that
λ = 2.068*
m