Question

Question 2.57 Dual behaviour of matter proposed by de Broglie led to the discovery of electron microscope often used for the highly magnified images of biological molecules and other type of material. If the velocity of the electron in this microscope is 1.6 × 106 ms–1, calculate de Broglie wavelength associated with this electron.

Class XI Structure of Atom Page 68

Answer 1

if the velocity of the electron in the microscope is 1.6 × 10^6 m/s . we know, mass of electron is 9.11 × 10^-31 Kg.
hence, we have to use De - broglie's wavelength ,
λ = h/mv
= 6.626 × 10^34 Js/9.11 × 10^-31 Kg × 1.6 × 10^6 m/s
= 0.455 × 10^-9 m
= 4.55 × 10^-10 m = 455 pm

hence, wavelength is 455 pm

Answer 2

Answer:

445 pm

Explanation:

velocity of electron = 1.6 x $10^{6}$ m/s

mass of electron     = 9.11 x $10^{-31}$ kg

by De Broglie's equation

λ=h/mv

λ= $\frac{6.62 * 10^{43} }{9.1 * 10^{-31} * 1.6 * 10^{6} }$

 =0.45 x $10^{-9}$m

 =4.5 x $10^{-10}$

 = 445 pm

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