Atoms Test

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Atoms Test - 14

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Weekly Quiz Competition

Question 1
1 / -0

Among the following, identify one that is antiparticle of itself :

A
Photon

B
Electron

C
Proton

D
$$\pi $$ -meson

Solution

  Particle   Anti particle
Electron Positron
Proton    Anti Proton
Neutron Anti Neutron
   Positive $$\pi$$ meson    Negative $$\pi$$ meson
  Positive $$\mu$$ meson    Negative $$\mu$$ meson
So, photon is its own anti particle.
Question 2
1 / -0

The particle which has no antiparticle is

A
Proton

B
Electron

C
Neutron

D
Photon

Solution

  Particle   Anti particle
Electron Positron
Proton    Anti Proton
Neutron Anti Neutron
   Positive $$\pi$$ meson    Negative $$\pi$$ meson
  Positive  $$\mu$$ meson    Negative $$\mu$$ meson
So, photon has no anti particle but Proton, Electron and Neutron has anti particle as given above.
Question 3
1 / -0

If the radius of first Bohr orbit is $$'r'$$, then radius of second orbit will be

A
$$2r$$

B
$$\dfrac{r}{2}$$

C
$$4r$$

D
$$\sqrt{2r}$$

Solution

$$r = \dfrac{E_0n^2h^2}{Rmze^2}$$

So, $$r \propto n^2$$

So, $$\dfrac{r_1}{r_2} = (\dfrac {n_1}{n_2})^2$$

$$\Rightarrow \dfrac{r}{r_2} = (\dfrac{n}{2n})^2$$ ($$\because$$second orbit has n=2)

$$\Rightarrow r_2 = 4r$$
Question 4
1 / -0

Cathode ray tube is operating at 5 kV. Then ,what is the K.E. acquired by the electrons ?

A
5 eV

B
5 MeV

C
5 keV

D
5 V

Solution

K.E.= p.d. x e
= 5kV x e
= 5 keV.
Question 5
1 / -0

According to Bohr hypothesis, discrete quantity is

A
Momentum

B
Angular velocity

C
Potential energy

D
Angular momentum

Solution

Bohr's hypothesis:
1) Electrons revolves round the nucleus in water orbits.
2) Orbit of the electron around the nucleus can take only some special values of radius.
3) The energy of the atom as a definite value in these orbits.
4) In this Orbits, Angular momentum (e) of the electron is integral multiple of the plank's constant h divided by $$2n$$ i.e. $$l = n \dfrac{h}{2n}$$.
Question 6
1 / -0

As the orbit number increases, the distance between two consecutive orbits in an atom or ion having single electron

A
increases

B
decreases

C
remains the same

D
first increases and then becomes constant

Solution

We know, $$r \propto \dfrac{n^{2}}{Z}$$
Distance between two consecutive orbits $$\propto n^2-(n-1)^2$$ = $$2n-1$$
So as n>0, when the orbit number increases, the distance between two consecutive orbits in an atom or ion having a single electron, increases.
Question 7
1 / -0

Bohrs atomic model assumes :

A
the nucleus is of infinite mass and is at rest

B
electron in a quantized orbit will not radiate energy

C
mass of the electron remains constant

D
all of the above

Solution

Bohr's Assumption :
( i ) Electron in atoms orbit the nucleus.
( ii ) Electron in a quantized orbit will not radiate energy.
( iii ) Electrons can only gain and lose energy by jumping from one allowed orbit to other.
He also assumed one of the postulates of Rutherford's atomic model, that is, the mass of the nucleus is very large compared to that of the electrons and hence assumed to be infinite
Question 8
1 / -0

For electron moving in $$n$$ th orbit of the atom, the angular velocity is proportional to:

A
$$n$$

B
$$\dfrac{1}{n}$$

C
$$n^{3}$$

D
$$\dfrac{1}{n^{3}}$$

Solution

Radius of nth orbit $$r_n = 0.529n^2$$ $$A^o$$ $$\implies r_n \propto n^2$$
Velocity of electron in nth orbit $$v_n = 2.18\times 10^6\dfrac{1}{n}$$ m/s $$\implies v_n \propto \dfrac{1}{n}$$
$$\therefore$$ Angular velocity $$w_n = \dfrac{v_n}{r_n} \propto \dfrac{1/n}{n^2}$$
$$\implies$$ $$w_n \propto \dfrac{1}{n^3}$$
Question 9
1 / -0

The incorrect statement from the following is:

A
Material wave (de-Broglie wave) can travel in vacuum.

B
Electromagnetic wave can travel through vacuum.

C
The velocity of photon is the same as light passes through any medium.

D
Wavelength of de-Broglie wave depends upon velocity.

Solution

The speed at which light propagates through medium depends on the refractive index of the material (n). It is given by:
$$v=c/n$$.
Question 10
1 / -0

The visible region of hydrogen spectrum was first studied by

A
Lyman

B
Balmer

C
Pfund

D
Brackett

Solution

Visible region of hydrogen spectrum was first studied by Balmer, therefore they are also called as Balmer lines.

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Atoms Test - 14

Result

Atoms Test - 14

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SHARING IS CARING

Question 1

Photon

Electron

Proton

$$\pi $$ -meson

Solution

Question 2

Proton

Electron

Neutron

Photon

Solution

Question 3

$$2r$$

$$\dfrac{r}{2}$$

$$4r$$

$$\sqrt{2r}$$

Solution

Question 4

5 eV

5 MeV

5 keV

5 V

Solution

Question 5

Momentum

Angular velocity

Potential energy

Angular momentum

Solution

Question 6

increases

decreases

remains the same

first increases and then becomes constant

Solution

Question 7

the nucleus is of infinite mass and is at rest

electron in a quantized orbit will not radiate energy

mass of the electron remains constant

all of the above

Solution

Question 8

$$n$$

$$\dfrac{1}{n}$$

$$n^{3}$$

$$\dfrac{1}{n^{3}}$$

Solution

Question 9

Material wave (de-Broglie wave) can travel in vacuum.

Electromagnetic wave can travel through vacuum.

The velocity of photon is the same as light passes through any medium.

Wavelength of de-Broglie wave depends upon velocity.

Solution

Question 10

Lyman

Balmer

Pfund

Brackett

Solution

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