A golf-ball weigh 40.0 g . If it is moving with a velocity of 20.0 ~ms^ -1 , it's de-Broglie wavelength is

Correct answer: C. 8.28 × 10^ -25 ~nm From de-Broglie equation, λ=(h)/(m v) aligned λ & = 6.626 × 10^ -34 j . s 40 × 10^ -3 ~kg × 20.0 ~ms^ -1 & =8.28 × 10^ -34 ~m aligned aligned & =8.28 × 10^ -34 × 10^9 ~nm & =8.28 × 10^ -25 ~nm aligned

JIPMER 2004

Exam: JIPMER
Year: 2004
Subject: Chemistry
Chapter: Structure of Atom

JIPMER 2004 · Chemistry · Structure of Atom

A golf-ball weigh 40.0 g . If it is moving with a velocity of 20.0...

A golf-ball weigh 40.0 g . If it is moving with a velocity of

20.0 \mathrm{~ms}^{-1}

, it's de-Broglie wavelength is

Options

A $1.66 \times 10^{-34} \mathrm{~nm}$
B $8.28 \times 10^{-32} \mathrm{~nm}$
C $8.28 \times 10^{-25} \mathrm{~nm}$
D $1.66 \times 10^{-24} \mathrm{~nm}$

Correct answer

C. $8.28 \times 10^{-25} \mathrm{~nm}$

Step-by-step solution

From de-Broglie equation,

\lambda=\frac{h}{m v}

\begin{aligned} \therefore \quad \lambda & =\frac{6.626 \times 10^{-34} \mathrm{j} . \mathrm{s}}{40 \times 10^{-3} \mathrm{~kg} \times 20.0 \mathrm{~ms}^{-1}} \\ & =8.28 \times 10^{-34} \mathrm{~m}\end{aligned}

\begin{aligned} & =8.28 \times 10^{-34} \times 10^9 \mathrm{~nm} \\ & =8.28 \times 10^{-25} \mathrm{~nm}\end{aligned}

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