Atomic Structure Test

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Atomic Structure Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

A photon has energy of 2.6 eV. What is its wavelength?

A
2.6 × 10^-7 m.

B
4.8 × 10^-7 m.

C
7.2 × 10^-7 m.

D
6.3 × 10^-7m

Solution

Given, 2.6 e V = 2.6 X 1.6 X 10^-19 J = 4.16 X 10^-19 J
Now,by using formula,
Question 2
1 / -0

If uncertainty in the position of an electron is zero, the uncertainity in its momentum would be

A
Zero

B
< h / 4π

C
Infinite

D
≤ h / 4π

Solution

Actually, according to uncertainty principle,
Δx.Δp ≥ h/4π
or Δx.Δp ≈ h/4n
where, Δx, is uncertainty in position and Δp is uncertainty in momentum.
So, if Δx = 0
Hence, uncertainty in momentum Δp = h/4π x 0 = Infinity.
Question 3
1 / -0

Uncertainty in the position of an electron (mass 9.1 x 10^-31 kg) moving with a velocity of 300 ms^-1, accurate upto 0.001% will be:

A
1.92 x 10^-2 m

B
3.84 x 10^-2 m

C
19.2 x 10^-2 m

D
5.76 x 10^-2 m

Solution

Change in position (X),
P= momentum,
P=mv.
X ×P = h/4π.
X× m v = h/4π.
X= h/4πmv.
X=6.625×10^-34/4 ×3.14×9.1×10^-31×300.
X=6.625×10^-5/342.8.
X=0.0192×10^-5.
X= 19.2× 10^-7.
in question it is given to take accurate up to 0.001%.= 1×10^-3/100=1 ×10^-5.
X= 19.2 ×10^-7/10^-5.
X=19.2 × 10^-2.
Question 4
1 / -0

What experimental result did Albert Einstein cite as evidence that light has particle-like properties?

A
The fact that electrons posses both particle and wave-like properties.

B
The orbits of electrons within the hydrogen atom could have only certain radii.

C
The ejection of electrons from a metal surface by light depends only on the frequency of the light used.

D
The pattern of light and dark bands produced when light was passed through a pair of narrow slits.

Solution

Einstein's explanation of photoelectric effect using the concept of photons. In 1905, Albert Einstein provided an explanation of the photoelectric effect, an experiment that the wave theory of light failed to explain. He did so by postulating the existence of photons, quanta of light energy with particulate qualities.
Question 5
1 / -0

The de Broglie wavelength of a particle is given by.

A
h/mv

B
h + mv

C
hmv

D
mv/c

Solution

According to De Broglie, a moving small particle like electron, proton, neutron, dust particle, a small ball etc. has the properties of a wave (Dual nature of matter).
The wave length of a moving particle may be calculated as
λ=h/mv =h/p
De Broglie equation
λ =wave length
h= planck’s constt.
m=mass of particle
v=velocity of particle
p=momentum of the particle
Question 6
1 / -0

The de – Broglie wavelength of an electron is 600 nm. The velocity of the electron having the mass 9.1 X 10^-31 Kg is

A
0.0012 x 10⁺⁴ m/s

B
0.0012 x 10⁺³ m/s

C
0.0012 x 10⁺⁶ m/s

D
0.0012 x 10⁺² m/s

Solution

1nm= 10^-9m.
wavelength= 600nm= 600 ×10-9m
wavelength= h/p.
wavelength= h/mv
v= h/ m wavelength
v=6.625×10^-34/9.1×10^-31×600× 10^-9.
v= 6.625 × 10⁺⁴/9.1 ×6.
v= 6.625× 10⁺⁴/9.1×6.
v= 6.625 ×10⁴/54.6.
v= 0.12 × 10⁴.
v= 0.0012 × 10⁺⁶ .
Question 7
1 / -0

In the relationship ∆x. ∆p = , ∆p is:

A
Certainty in momentum

B
Certainty in position

C
Uncertainty in momentum

D
Uncertainty in position

Solution

The uncertainty principle is alternatively expressed in terms of a particle’s momentum and position. The momentum of a particle is equal to the product of its mass times its velocity. Thus, the product of the uncertainties in the momentum and the position of a particle equals h/(4π)
Question 8
1 / -0

The uncertainty principle states that

A
it is impossible to know anything with certainty

B
there can only be one uncertain digit in a reported number

C
it is impossible to know the exact position and momentum of an electron simultaneously

D
it is impossible to determine mass of light particles

Solution

The Heisenberg uncertainty principle states that it is impossible to know simultaneously the exact position and momentum of a particle. That is, the more exactly the position is determined, the less known the momentum, and vice versa.
Question 9
1 / -0

Whch of the following will have the maximum value of ∆v . ∆x

A
Cricket ball

B
Electron

C
Table

D
Tennis ball

Solution

This is accordance with the Heisenberg's uncertainty principle.It states that it is not possible to accurately determine the position and momentum of a microscopic particle such as electron simultaneously.
Question 10
1 / -0

The de Broglie wavelength of an electron is 8.7 x 10^-11 m. The mass of an electron is 9.1 x 10^-31 kg. The velocity of this electron is:

A
0.0837 × 10⁸ m/s

B
6.9 × 10^-5 m/s

C
8.4 × 10³ m/s

D
1.2 × 10^-7 m/s

Solution

Wavelength = h/p = h/ mv.
v= h/ m.wavelength
v= 6.625 × 10^-34/9.1×10^-31 × 8.7 × 10^-11
v= 6.625 × 10⁸/ 9.1× 8.7
v= 0.08367 ×10⁸.

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Atomic Structure Test - 21

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Question 1

2.6 × 10-7 m.

4.8 × 10-7 m.

7.2 × 10-7 m.

6.3 × 10-7m

Solution

Question 2

Zero

< h / 4π

Infinite

≤ h / 4π

Solution

Question 3

1.92 x 10-2 m

3.84 x 10-2 m

19.2 x 10-2 m

5.76 x 10-2 m

Solution

Question 4

The fact that electrons posses both particle and wave-like properties.

The orbits of electrons within the hydrogen atom could have only certain radii.

The ejection of electrons from a metal surface by light depends only on the frequency of the light used.

The pattern of light and dark bands produced when light was passed through a pair of narrow slits.

Solution

Question 5

h/mv

h + mv

hmv

mv/c

Solution

Question 6

0.0012 x 10+4 m/s

0.0012 x 10+3 m/s

0.0012 x 10+6 m/s

0.0012 x 10+2 m/s

Solution

Question 7

Certainty in momentum

Certainty in position

Uncertainty in momentum

Uncertainty in position

Solution

Question 8

it is impossible to know anything with certainty

there can only be one uncertain digit in a reported number

it is impossible to know the exact position and momentum of an electron simultaneously

it is impossible to determine mass of light particles

Solution

Question 9

Cricket ball

Electron

Table

Tennis ball

Solution

Question 10

0.0837 × 108 m/s

6.9 × 10-5 m/s

8.4 × 103 m/s

1.2 × 10-7 m/s

Solution

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2.6 × 10^-7 m.

4.8 × 10^-7 m.

7.2 × 10^-7 m.

6.3 × 10^-7m

1.92 x 10^-2 m

3.84 x 10^-2 m

19.2 x 10^-2 m

5.76 x 10^-2 m

0.0012 x 10⁺⁴ m/s

0.0012 x 10⁺³ m/s

0.0012 x 10⁺⁶ m/s

0.0012 x 10⁺² m/s

0.0837 × 10⁸ m/s

6.9 × 10^-5 m/s

8.4 × 10³ m/s

1.2 × 10^-7 m/s