Atoms Test

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Atoms Test - 28

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Weekly Quiz Competition

Question 1
1 / -0

In the hydrogen atom in the ground state

A
the kinetic energy of the electron is less than the potential energy which is positive

B
the potential energy is less than the kinetic energy which is positive

C
the potential energy is negative and the kinetic energy numerically less than the numerical value of potential energy

D
the total energy is negative

Solution

$$ \begin{array}{l} \text { we know that total energy } \\ \text { of an electron in } n^{\text {th }} \text { orbit } \\ \text { is }=-13.6 \frac{z^{2}}{n^{2}} ev \\ \text { Total Energy }=-13.6 \frac{z^{2}}{n^{2}} e v \end{array} $$
$$ \begin{array}{l} \text { Total Energy= Kinetic Energy+Potential energy } \\ \qquad \begin{array}{l}T \cdot E= K \cdot E +P \cdot E \end{array} \\ \text { & } K \cdot E = - \frac{P \cdot E}{2} \end{array} $$
$$ \begin{aligned} \therefore T \cdot E =\frac{-P \cdot E \cdot}{2}+P \cdot E \cdot \\ \frac{-13 \cdot 6 z^{2}}{n^{2}} e v &=\frac{P \cdot E}{2} \end{aligned} $$
$$ \begin{aligned} \therefore P \cdot E \cdot &=-\frac{2 \times 13 \cdot 6 \times z^{2}}{n^{2}} \\ &=-27 \cdot 2 \times \frac{z^{2}}{n^{2}} \end{aligned} $$
$$ \begin{aligned} \text { & } K \cdot E \cdot &=-\frac{P \cdot E}{2} \\ &=13 \cdot 6 \frac{z^{2}}{n^{2}} e v \\ & \text { for hydrogen's ground state } \\ z=1, & n=1 \end{aligned} $$
$$ \begin{aligned} P \cdot E =& \frac{-27 \cdot 2 \times (1)^{2}}{1} \\ =&-27. 2 e v \\ \& K.E= \frac{13.6 \times (1)^2}{1} = 13.6ev \\ \text { Hence, option } c \text { is correct. } \end{aligned} $$
Question 2
1 / -0

The radius of shortest orbit in one electron system is $$18 $$pm. It may be

A
$$\displaystyle_{1}^{1}\textrm{H}$$

B
$$\displaystyle_{1}^{2}\textrm{H}$$

C
$$\displaystyle He^{+}$$

D
$$\displaystyle Li^{++}$$

Solution
Question 3
1 / -0

A cathode ray tube has a potential difference of V volt between the cathode and the anode. The speed of cathode rays is given by

A
$$\displaystyle 6\times 10^{-5}\sqrt{V}cm/s$$

B
$$\displaystyle 6\times 10^{5}\sqrt{V}m/s$$

C
$$\displaystyle 6\times 10^{7}\sqrt{V}cm/s$$

D
$$\displaystyle 6\times 10^{3}\sqrt{V}m/s$$

Solution

The kinetic energy of electrons is
$$E = \dfrac{1}{2} mv^2$$
The potential between cathode and anode is V hence,
$$\dfrac{1}{2} mv^2 = eV$$
$$v^2 = \dfrac{2eV}{m}$$
$$v^2 = \dfrac{2 \times 1.6 \times 10^{-19} V}{9.1 \times 10^{-31}}$$
$$v^2 = 0.35 \times 10^{12} V$$
$$v = \sqrt {35 \times 10^{10} V}$$
$$v = 5.9 \times 10^5 \sqrt V$$
$$v \approx 6 \times 10^5 \sqrt V ms^{-1}$$
Question 4
1 / -0

The collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate. Light source is made on and a saturation photocurrent is recorded. An electric field is switched on which has a vertically downward direction.

A
The photocurrent will increase

B
The kinetic energy of the electrons will increase

C
The stopping potential will decrease

D
The threshold wavelength will increase

Solution

Force on a electron in a electric field is given by:
$$F=eE$$.
So, if an electric field is switched on which has a vertically downward direction then the kinetic energy of the electrons will increase and will reach the collector plate in an experiment on photoelectric effect is kept vertically above the emitter plate.
So, the answer is option (B).
Question 5
1 / -0

The current in a photoelectric tube

A
Decreases with increase in frequency of the incident photon

B
Increases with increase in frequency of the incident photon

C
Decreases with increases in intensity of illumination

D
Increases with increase in intensity of illumination

Solution

$$ \begin{array}{l} \text { As, the intensity of light } \\ \text { increases, current increases, } \\ \text { this is because a larger number } \\ \text { of photons now fall on the } \\ \text { metal surface and hence a } \end{array} $$
$$ \begin{array}{l} \text { larger number of electrons } \\ \text { interact with photons. The } \\ \text { number of electrons emitted } \\ \text { increases and hence the current } \\ \text { increases. } \end{array} $$
$$ \begin{array}{l} \text { Hence the current in photoelectric } \\ \text { tube increases with increase } \\ \text { in intensity of illumination. } \end{array} $$
Question 6
1 / -0

Which of the following products in a hydrogen atom is independent of principal quantum number $$n$$ ?

A
$$ v_r $$

B
$$ v_n $$

C
$$ E_r^{2}$$

D
$$ E_n$$

Solution

The velocity of revolving electron is:
$$v = \dfrac{e^2}{2nh\epsilon_0}$$

Hence, $$v_n = \dfrac{e^2}{2nh\epsilon_0}(n)$$
$$v_n = \dfrac{e^2}{2h\epsilon_0}$$

The product thus obtained does not contain principal quantum number(n) hence $$v_n$$ is independent of $$n$$.
Question 7
1 / -0

When a photomultiplier tube was used , the photo current recorded is 60 $$\mu A$$. The actual photo current is

A
$$\displaystyle > 60$$ $$ \mu A$$

B
$$\displaystyle = 60$$ $$ \mu A$$

C
$$\displaystyle < 60$$ $$ \mu A$$

D
None of these

Solution

The current has to be $$< 60 \mu A$$.
Since dark current is also involved in the measured current value,
$$ I_{actual} = I_{measured} - I_{dark}$$
Question 8
1 / -0

If the spins of electrons of atoms in a substance are paired, then the substance will have

A
Paramagnetic Nature

B
Ferro magnetic Nature

C
Diamagnetic Nature

D
None of these

Solution

$$\begin{array}{l}\text { Spins are paired means clockwise and anticlock } \\\text { wise rotation cancel each other, so it does not have } \\\text { any free electron, } \\\therefore\text { it is in dimagnetic in nature. }\end{array}$$
Question 9
1 / -0

Cathode ray oscillograph is used for

A
Taking X-ray photographs

B
Showing pictures

C
Demonstrating electrical pulses

D
Producing elementary particles

Solution

$$\begin{array}{l}\text { The cathode ray oscilloscope (CRO) } \\\text { is a common laboratory instrument } \\\text { that provides accurate time } \\\text { and aplitude measurements of } \\\text { voltage signals over a wide } \\\text { range of frequencies. Its } \\\text { reliability, stability and ease of } \\\text { opteration make it suitable } \\\text { as a general purpose laboratory } \\\text { instrument. } \\\text { In easy words, it is used } \\\text { for demonstrating } \\\text { electrical pulses. }\end{array}$$
Question 10
1 / -0

Raman effect shows

A
polarisation

B
quantum nature

C
Brillouin zone scattering

D
wave nature of light

E
none of the above

Solution

Raman effect deals with inelastic scattering of photon particles by interaction with vibrational and rotational transitions in the bonds( Raman scattering can occur with a change in energy of a molecule due to a transition) it has nothing to do with polarisation , quantum nature , wave nature (photon scattering ) and Brillouin zone scattering
option $$E$$ is correct

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Atoms Test - 28

Result

Atoms Test - 28

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Question 1

the kinetic energy of the electron is less than the potential energy which is positive

the potential energy is less than the kinetic energy which is positive

the potential energy is negative and the kinetic energy numerically less than the numerical value of potential energy

the total energy is negative

Solution

Question 2

$$\displaystyle_{1}^{1}\textrm{H}$$

$$\displaystyle_{1}^{2}\textrm{H}$$

$$\displaystyle He^{+}$$

$$\displaystyle Li^{++}$$

Solution

Question 3

$$\displaystyle 6\times 10^{-5}\sqrt{V}cm/s$$

$$\displaystyle 6\times 10^{5}\sqrt{V}m/s$$

$$\displaystyle 6\times 10^{7}\sqrt{V}cm/s$$

$$\displaystyle 6\times 10^{3}\sqrt{V}m/s$$

Solution

Question 4

The photocurrent will increase

The kinetic energy of the electrons will increase

The stopping potential will decrease

The threshold wavelength will increase

Solution

Question 5

Decreases with increase in frequency of the incident photon

Increases with increase in frequency of the incident photon

Decreases with increases in intensity of illumination

Increases with increase in intensity of illumination

Solution

Question 6

$$ v_r $$

$$ v_n $$

$$ E_r^{2}$$

$$ E_n$$

Solution

Question 7

$$\displaystyle > 60$$ $$ \mu A$$

$$\displaystyle = 60$$ $$ \mu A$$

$$\displaystyle < 60$$ $$ \mu A$$

None of these

Solution

Question 8

Paramagnetic Nature

Ferro magnetic Nature

Diamagnetic Nature

None of these

Solution

Question 9

Taking X-ray photographs

Showing pictures

Demonstrating electrical pulses

Producing elementary particles

Solution

Question 10

polarisation

quantum nature

Brillouin zone scattering

wave nature of light

none of the above

Solution

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