Question

what accelerating potential is needed to produce electron beam which wave length of 9picometre.
give full explanation.​

Answer 1

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We assume that momentum (p) and mass (m) are required to find the potential difference (V). Therefore,

p²/2m= eV = gain in kinetic energy

where, P = √2meV

According to De Broglie's equation,

Gamma = h/p

Gamma =h/√2meV

or,

where,

m = mass of an electron = 9.1×10-³¹ kg

e = charge on an electron = 1.6×10-¹⁹

wavelength = 0.09 A = 9×10-¹⁰

h = plank's conctant = 6.626×10-³⁴

Therefore, putting the given values into the above formula as follows.

$9 \times {10}^{ - 10} = \frac{6.626 \times {10}^{ - 34} }{2 \times 9.1 \times 10 ^{ - 31} \times 1.6 \times 16 ^{ - 19} \times v}$

V = $0.0252 \times 10 ^{74}$

Therefore, we can conclude that amount of accelerating potential is $0.0252 \times 10 ^{74} v$

Answer 2

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