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11th Standard English Medium Chemistry Subject Quantum Mechanical Model of Atom Book Back 5 Mark Questions with Solution Part - II

Published on: 08/06/2021

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Std: 11th Standard - Chemistry

chapter: - Book back Questions

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Questions:

Generic 3

3 x 3 = 9

Show that if the measurement of the uncertainty in the location of the particle is equal to its de Broglie wavelength, the minimum uncertainty in its velocity is equal to its velocity 1/4\(\pi\) of its velocity (V).

Protons can be accelerated in particle accelerators. Calculate the wavelength (in Å) of such accelerated proton moving at 2.85 x 10⁸ ms^-1 (the mass of proton is 1.673 x 10^-27 Kg).

The Li²⁺ ion is a hydrogen like ion that can be described by the Bohr model. Calculate the Bohr radius of the third orbit and calculate the energy of an electron in 4^th orbit.

Generic 5

2 x 5 = 10

Identify the missing quantum numbers and the sub energy level

n	1	m	Sub energy level
?	?	0	4d
3	1	0	?
?	?	?	5p
?	?	-2	3d

What is the de Broglie wavelength (in cm) of a 160 g cricket ball travelling at 140 Km hr ^-1.

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Answers Key:

Generic 3

3 x 3 = 9

\(\triangle x=?\)
\(\triangle v=?\)
\(\triangle x.\triangle p\ge\frac{h}{4\pi}\)
\(\lambda (m\triangle v)\ge\frac{h}{4\pi}\)
\(\triangle v\ge\frac{h}{4\pi(m\lambda)}\)
\(\triangle \ge\frac{h}{4\pi\times m \times\frac{h}{mv}}\) \(\left[ \because \lambda-={{h}\over{mv}} \right]\)
\(\triangle v \ge \frac{v}{4\pi}\)
therefore, minimum uncertainty in velocity \(=\frac{v}{4\pi}\)

v = 2.85 x 10⁸ms^-1
m_p = 1.673 x 10^-27Kg
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times10^{-34}kgm^{2}s^{-1}}{1.673\times10^{-27}kg\times\times2.85\times10^{8}ms^{-1}}\)
\(\lambda=1.389\times10^{-15}\Rightarrow\lambda=1.389\times10^{-15}A\) \([\because \mathring{A}={10}^{-10}m]\)

\({ r_{n} }=\frac{(0.529)n^2}{z}\mathring{A}\ \ { E_{n} }=\frac{-13.6(z)^2}{(n)^2}ev atom^{-1}\)
for Li²⁺ z = 3
Bohr radius for the third orbit (r₃)
= \(\frac { (0.529){ (3) }^{ 2 } }{ 3 } \)
= 0.529\(\times\)3
=1.587 \(\mathring{A}\)
Energy of an electron in the fourth orbit
\(({E}_{6})=\frac { -13.6{ (3) }^{ 2 } }{ { (4) }^{ 2 } } \)
=-7.65eV atom^-1

Generic 5

2 x 5 = 10

n	1	m	Sub energy level
4	2	0	4d
3	1	0	3p
5	1	any one value -1,0,+1	5p
3	2	-2	3d

m 160 g = 160 x 10^-3 kg
\(v=140 km hr^{-1}=\frac{104\times10^{3}}{60\times60}ms^{-1}\)
\(v=38.88ms^{-1}\)
\(\lambda=\frac{h}{mv}\)
\(=\frac{6.626\times^{-34}Kgm^{2}s^{-1}}{160\times10^{-3}Kg\times38.88ms^{-1}}\)
\(\lambda=1.065\times10^{-34}m\)