Question

An electron having initial K.E of 6ev is accelerated through P.D. of4V. Find the wavelength associated with electron.
[Ans=3.87x10^-10]

Answer 1

Explanation:

We have initial kinetic energy as 6eV plus the potential difference as 4V.

Hence, the total kinetic energy = 6+4 eV

                                                    = 10 eV = 10 x 1.6 x 10^{-19} J

                                                    = 16 x 10^{-19} J

1/2 mv2 = 16 x 10-19 J

∴ v = √3.4 x 10^12 m/s

Hence, by using De broglie equation:

λ = h/mv

λ = (6.63 x 10-34) / (9.1 x 10-31) (√3.4 x 10^12)

∴ λ = 3.87 x 10^-10

Answer 2

Explanation:

$\sqrt{15}$

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