Chapter 3, Problem 3.67QA

Interpretation Introduction

To find:

The wavelengths of the given objects.

Answer to Problem 3.67QA

Solution:

a) 10.8 nm

b) 0.180 nm

c) 8.2 x 10^-28 nm

d) 3.7 x 10^-54 nm

Explanation of Solution

1) Concept:

According to De Broglie,

All material particles in motion possess wave characteristics.

So, according to him, the wave-length associated with a particle of mass ‘m’ and moving with a velocity ‘v’ is given by the relation,

λ = $\frac{h}{mv}$ where ‘h’ is Planck’s constant.

2) Formula:

λ = $\frac{h}{mv}$

3) Calculations:

(a)

Mass of muon = 1.884 x 10^-28 kg,

velocity = 325 m/s,

h (Planck’s Constant) = 6.626 x 10^-34 J.s.

So wavelength (λ) = $\frac{6.626 \times {10}^{-34 } J.s}{1.884 \times {10}^{-28 }kg \times 325 .m/s}$ x $\frac{1 kg.m^{2 }/s^2}{1J}=0.0108 x {10}^{-6 }m$

$0.0108\mathrm{ }\mathrm{x}\mathrm{ }{10}^{-6\mathrm{ }}\mathrm{m}\mathrm{ }\mathrm{x}\mathrm{ }\frac{{10}^9 nm}{1 m}= \mathrm{ }10.8\mathrm{ }\mathrm{n}\mathrm{m}$

(b)

Mass of electron = 9.10938 x 10^-31 kg,

velocity = 4.05 x 10⁶ m/s,

h (Planck’s Constant) = 6.626 x 10^-34 J.

So wavelength (λ) = $\frac{6.626 \times {10}^{-34 } J}{9.10938 \times {10}^{-31} kg \times 4.05 \times {10}^6 m/s}$  x $\frac{1 kg.m^{2 }/s^2}{1J}$ = 0.180 x 10^-9m

$0.180\mathrm{ }\mathrm{x}\mathrm{ }{10}^{-9\mathrm{ }}\mathrm{m}\mathrm{ }\mathrm{x}\mathrm{ }\frac{{10}^9 nm}{1 m}= \mathrm{ }0.180\mathrm{ }\mathrm{n}\mathrm{m}$

(c)

Mass of sprinter = 82 kg,

velocity =9.9 m/s,

h (Planck’s Constant) = 6.626 x 10^-34 J.s

So wavelength (λ) = $\frac{6.626 \times {10}^{-34 } J}{82 \times 9.9 kg .m/s}$  x $\frac{1 kg.m^{2 }/s^2}{1J}$ = 0.0082 x 10^-34 m

$0.0082\mathrm{ }\mathrm{x}\mathrm{ }{10}^{-34\mathrm{ }}\mathrm{m}\mathrm{ }\mathrm{x}\mathrm{ }\frac{{10}^9 nm}{1 m}= \mathrm{ }8.2\mathrm{ }\mathrm{x}\mathrm{ }{10}^{-28}\mathrm{ }\mathrm{n}\mathrm{m}$

(d)

Mass of earth =6 x 10²⁴ kg,

velocity =3 x 10⁴ m/s,

h (Planck’s Constant) = 6.626 x 10^-34 J.s

So wavelength (λ) = $\frac{6.626 \times {10}^{-34 } J}{6 \times {10}^{24} \times 3 \times {10}^4 kg .m/s}$  x $\frac{1 kg.m^{2 }/s^2}{1J}$  = 0.37 x 10^-62 m

$0.37\mathrm{ }\mathrm{x}\mathrm{ }{10}^{-62\mathrm{ }}\mathrm{m}\mathrm{ }\mathrm{x}\mathrm{ }\frac{{10}^9 nm}{1 m}= \mathrm{ }3.7\mathrm{ }\mathrm{x}\mathrm{ }{10}^{-54}\mathrm{ }\mathrm{n}\mathrm{m}$

Conclusion:

Therefore the respective wavelengths for (a),(b),(c) and (d) are 10.8 nm, 0.18 nm, 8.2 x 10^-28 nm and 3.7 x 10^-54 nm.