The de Broglie wavelength of an electron moving with constant velocity is
0.367 nm . The mass of proton is 1835 times that of an electron . The de Broglie wavelength of a proton moving with the same velocity will be
A)0.2\10^12 m
B)0.3\10^12 m
C)0.4\10^12m
D)0.5\10^12 m
Answers
Answered by
7
(A) is answer
0.367×10^-9 =lambda(e-)=h/M(e-)v -----》(1)
lambda(p+)=h/M(p+)v 1835×Me-=Mp+
lambda(p+)=h/1835×M(e-)v
from (1)
lambda(p+)=0.367×10^-9/1835=0.2/10^12
0.367×10^-9 =lambda(e-)=h/M(e-)v -----》(1)
lambda(p+)=h/M(p+)v 1835×Me-=Mp+
lambda(p+)=h/1835×M(e-)v
from (1)
lambda(p+)=0.367×10^-9/1835=0.2/10^12
Similar questions