Atoms Test

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Atoms Test - 33

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Question 1
1 / -0

A beam of $$13.0\space eV$$ electrons is used to bombard gaseous hydrogen. The series obtained in emission spectra is/are

A
Lyman series

B
Balmer series

C
Brackett series

D
All of these

Solution

For $$13.0\ eV$$ we can expect emissions from any of the lines, Lyman, Balmer, Paschen , Brackett, Pfund, since all their energies lie within $$13\ eV$$.
Question 2
1 / -0

Which of the following is true when Bohr gave his model for hydrogen atom?

A
It was not known that hydrogen lines could be explained as differences of terms like $$R/n^2$$ with $$R$$ being a constant and $$n$$ an integer

B
It was not known that positive charge is concentrated in a nucleus of small size

C
It was known that radiant energy occurred in energy bundles defined by $$hv$$ with $$h$$ being a constant and $$v$$ a frequency

D
Bohr knew terms like $$R/n^2$$ and in the process of choosing allowed the orbits to fit them, he got angular momentum = $$nh/2\pi$$ as deduction

Solution

Niels Bohr introduced the atomic Hydrogen model in 1913. He described it as a positively charged nucleus, comprised of protons and neutrons, surrounded by a negatively charged electron cloud. In the model, electrons orbit the nucleus in atomic shells. The atom is held together by electrostatic forces between the positive nucleus and negative surroundings.
Bohr’s Postulates
1. An atom has a number of stable orbits in which an electron can reside without the emission of radiant energy. Each orbit corresponds, to a certain energy level.
2. An electron may jump spontaneously from one orbit (energy level E1) to the other orbit (energy level E2) (E2 > E1); then the energy change AE in the electron jump is given by Planck’s equation
∆E = E2-E1 = hv
Where h = Planck’s constant.
And v = frequency of light emitted.
3. The motion of an electron in a circular orbit is restricted in such a manner that its angular momentum is an integral multiple of h/2π, Thus
mvr = nh/2π, where m = mass of the electron
v = velocity of the electron
r = radius of the orbit an
n = an integer called principal quantum number of the electron.:
4. A special surface around nucleus which contained orbits of equal energy and radius was called shell. These shells are numbered from inside to outwards as 1, 2, 3, 4 etc. and called K, L, M, N etc., respectively.
Question 3
1 / -0

Difference between $$nth$$ and $$(n+1)th$$ Bohr's radius of hydrogen atom is equal to $$(n-1)th$$ Bohr's radius. The value of $$n$$ is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

Given,
$${ \left( n+1 \right) }^{ 2 }-{ n }^{ 2 }={ \left( n-1 \right) }^{ 2 }\\ \Rightarrow 1+2n={ n }^{ 2 }+1-2n\\ \Rightarrow { n }^{ 2 }=4n\\ \Rightarrow n=4$$
Question 4
1 / -0

The radius of hydrogen atom in the ground state is $$5.3\times10^{-11} m$$. When struck by an electron, its radius is found to be $$21.2\times10^{-11}m$$. The principal quantum number of the final state will be

A
$$2$$

B
$$3$$

C
$$4$$

D
$$16$$

Solution

From the concept attached:
$$ r \propto n^2$$
For $$n=1$$ if $$r_1 =5.3 \times 10^{-11}m$$
Then, $$r = 21.4 \times 10^{-11} = 4r_1 = 2^2 r_1$$ so $$n =2$$
Question 5
1 / -0

If elements with principal quantum number $$n>4$$ were not allowed in nature, the number of possible elements would have been

A
$$32$$

B
$$60$$

C
$$18$$

D
$$4$$

Solution

$$\textbf{Hint}$$:The total number of possible electronic configurations will give total number of elements.

$$\textbf{Step1:Number of elements}$$
The number of electrons that can fill a shell is given by twice the square of the quantum number of that shell. The total number of possible electronic configurations will give the total number of elements.
$$N=2n^2$$
where N is the number of electrons in a particular orbit or number of possible electronic configurations and n is the quantum number of the orbit.

$$\textbf{Step2:Put values of n}$$
For first orbit n=1
$$N=2(2^1)$$=4

For second orbit n=2
$$N=2(2^2)$$=8

For third orbit n=3
$$N=2(2^3)$$=16

For fourth orbit n=4
$$N=2(2^4)$$=32

Hence the total number of electronic configurations are=$$4+8+16+32=60$$
So total number of elements are =60.
Question 6
1 / -0

The velocity of an electron in the first orbit of $$H$$ atom is $$'v\ '$$. The velocity of an electron in the second orbit of $$He^+$$ is

A
$$2v$$

B
$$v$$

C
$$\displaystyle\frac{v}{2}$$

D
$$\displaystyle\frac{v}{4}$$

Solution

For hydrogen:
Angular momentum $$=mvr=\dfrac{nh}{2\pi}$$
$$v=\dfrac{nh}{mr2\pi}$$
$$r=0.529\dfrac{n^2}{Z}$$

So, $$v=\dfrac{Zh}{0.529mn2\pi}$$

For hydrogen $$Z=1,n=1$$
$$v=\dfrac{h}{0.529m2\pi}$$

For $$He^+$$ $$Z=2 ,n=2$$
$$v_1=\dfrac{2h}{0.529m\times 2\times 2\pi}=v$$
Question 7
1 / -0

How can the rate of emission of electrons in a cathode ray tube can be increased?

A
Increasing resistance of filament

B
Increasing filament current

C
Decreasing filament current

D
Increasing magnetic field inside tube

Solution

Rate of emission of electrons can be increased by increasing the temperature and temperature of cathode can be increased by increasing the filament current.
Question 8
1 / -0

In a cathode ray tube, what is the effect on the beam of particles if the anode voltage is increased ?

A
Kinetic Energy of electron beam increase

B
Rate of electron emission increase

C
Kinetic energy of electron beam remains unchanged

D
Rate of electron emission decrease

Solution

Electrons emitted by the cathode are attracted to the focusing anode if the anode voltage is increased.
Thus, Kinetic Energy of electron beam increase.
Question 9
1 / -0

In a cathode ray tube, what is the effect on the beam of particles if a hotter filament is used ?

A
Rate of electron emission remains unchanged

B
Rate of electron emission decreases

C
Rate of electron emission increases

D
Rate of electron emission increases then decreases

Solution

Hotter filament will increase the temperature.
Hence rate of electron emission will increase.
Question 10
1 / -0

State the approximate potential difference applied between the anode and filament cathode ray tube .

A
1000 V

B
50 V

C
5 V

D
1 V

Solution

The electrons are accelerated by a second anode at high potential, more than 500 V.
So, the answer is option (A).

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Atoms Test - 33

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Atoms Test - 33

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Question 1

Lyman series

Balmer series

Brackett series

All of these

Solution

Question 2

It was not known that hydrogen lines could be explained as differences of terms like $$R/n^2$$ with $$R$$ being a constant and $$n$$ an integer

It was not known that positive charge is concentrated in a nucleus of small size

It was known that radiant energy occurred in energy bundles defined by $$hv$$ with $$h$$ being a constant and $$v$$ a frequency

Bohr knew terms like $$R/n^2$$ and in the process of choosing allowed the orbits to fit them, he got angular momentum = $$nh/2\pi$$ as deduction

Solution

Question 3

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 4

$$2$$

$$3$$

$$4$$

$$16$$

Solution

Question 5

$$32$$

$$60$$

$$18$$

$$4$$

Solution

Question 6

$$2v$$

$$v$$

$$\displaystyle\frac{v}{2}$$

$$\displaystyle\frac{v}{4}$$

Solution

Question 7

Increasing resistance of filament

Increasing filament current

Decreasing filament current

Increasing magnetic field inside tube

Solution

Question 8

Kinetic Energy of electron beam increase

Rate of electron emission increase

Kinetic energy of electron beam remains unchanged

Rate of electron emission decrease

Solution

Question 9

Rate of electron emission remains unchanged

Rate of electron emission decreases

Rate of electron emission increases

Rate of electron emission increases then decreases

Solution

Question 10

1000 V

50 V

5 V

1 V

Solution

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