An electron is accelerated by a potential difference of 50 Volt ., then the de-Broglie wavelength associated with it .
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De Broglie wavelength of a particle is given by
λ=h/p. (1)
where h is plank's constant and p is momentum of the particle.
λ=h/mv. (2)
For an electron accelerated with potential V
KE=1/2mv²=qV
where m v, V and q are mass velocity potential and charge of the particle respectively.
v=√(2qV/m)
putting the value of v in equation (2) we get
λ=h/(m√(2qV/m))
=h/√(2mqV). (3)
now h= 6.6 x 10^-34 m²kg/s
m=9.1 x10^-31 kg
q=-1.6x10^-19C
V=50V
so
λ=1.735x10^-10 m=1.735Angstrom
Hope it helps.
P.S.Next time use this direct formula (only for electrons)
h=12.27/√V Angstrom.You can derive this formula by putting all the values except V in equation (3)
λ=h/p. (1)
where h is plank's constant and p is momentum of the particle.
λ=h/mv. (2)
For an electron accelerated with potential V
KE=1/2mv²=qV
where m v, V and q are mass velocity potential and charge of the particle respectively.
v=√(2qV/m)
putting the value of v in equation (2) we get
λ=h/(m√(2qV/m))
=h/√(2mqV). (3)
now h= 6.6 x 10^-34 m²kg/s
m=9.1 x10^-31 kg
q=-1.6x10^-19C
V=50V
so
λ=1.735x10^-10 m=1.735Angstrom
Hope it helps.
P.S.Next time use this direct formula (only for electrons)
h=12.27/√V Angstrom.You can derive this formula by putting all the values except V in equation (3)
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