Atoms Test

Result

Atoms Test - 63

Score
-
out of -
Rank
-
out of -

TIME Taken - -

SHARING IS CARING

If our Website helped you a little, then kindly spread our voice using Social Networks. Spread our word to your readers, friends, teachers, students & all those close ones who deserve to know what you know now.

Weekly Quiz Competition

Question 1
1 / -0

According to de-Broglie wavelength for electron in an orbit of hydrogen atom is $$10^{-9}\ m$$. The principle quantum number for this electron is

A
$$1$$

B
$$2$$

C
$$3$$

D
$$4$$

Solution

As we know,
$$2\pi rn= \lambda $$

$$\Rightarrow n=\dfrac {\lambda}{2\pi r}=\dfrac {10^{-9}}{2\times 3.14 \times 5.13\times 10^{-11}}=3$$
Question 2
1 / -0

The spectral series of the hydrogen spectrum that lies in the ultraviolet region is the

A
Balmer series

B
Pfund series

C
Paschen series

D
Lyman series

Solution

Lyman series of hydrogen atom lies in ultraviolet region, Balmer series lies in visible region while found and paschen series lie in infrared region.
Question 3
1 / -0

According to Bohr's theory the expression for the kinetic and potential energy of an electron revolving in an orbit is given respectively by

A
$$+\dfrac{e^{2}}{8\pi\varepsilon_{0}r}$$ and $$-\dfrac{e^{2}}{4\pi\varepsilon_{0}r}$$

B
$$+\dfrac{8\pi\varepsilon_{0}r}{r}$$ and $$-\dfrac{4\pi\varepsilon_{0}r}{r}$$

C
$$-\dfrac{e^{2}}{8\pi\varepsilon_{0}r}$$ and $$-\dfrac{e^{2}}{4\pi\varepsilon_{0}r}$$

D
$$+\dfrac{e^{2}}{8\pi\varepsilon_{0}r}$$ and $$+\dfrac{e^{2}}{4\pi\varepsilon_{0}r}$$

Solution

$$P.E.=-\dfrac{ke^{2}}{r}=-\dfrac{e^{2}}{4\pi\varepsilon_{0}r};$$
$$K.E.=-\dfrac{1}{2}(P.E,)=\dfrac{e^{2}}{8\pi\varepsilon_{0}r}$$
Question 4
1 / -0

To explain his theory, Bohr-used

A
Conservation of linear momentum

B
Conservation of angular momentum

C
Conservation of quantum frequency

D
Conservation of energy

Solution

Bohr postulated that the angular momentum pf the electron is conserved.
Question 5
1 / -0

In the Bohr's hydrogen atom model, the radius of the stationary orbit is directly proportional to ($$n=$$ principle quantum number)

A
$$n^{-1}$$

B
$$n$$

C
$$n^{-2}$$

D
$$n^{2}$$

Solution

Bohr radius $$r=\dfrac{\varepsilon_{0}n^{2}h^{2}}{\pi Zme^{2}};$$
$$\therefore r\propto n^{2}$$
Question 6
1 / -0

Which one of these is non-divisible

A
Nucleous

B
Photon

C
Proton

D
Atom

Solution
Question 7
1 / -0

According to Bohr's theory the radius of electron in an orbit described by principle quantum number $$n$$ and atomic number $$Z$$ is proportional to

A
$$Z^{2}n^{2}$$

B
$$\dfrac{Z^{2}}{n^{2}}$$

C
$$\dfrac{Z^{2}}{n}$$

D
$$\dfrac{n^{2}}{Z}$$

Solution

$$r=\dfrac{\varepsilon_{0}n^{2}h^{2}}{\pi Zme^{2}};$$
$$ \therefore r\propto \dfrac{n^{2}}{Z}$$
Question 8
1 / -0

Which of the following is true.

A
Lyman series is a continuous spectrum

B
Paschen series is a line spectrum in the infrared

C
Balmer series is a line spectrum in the ultraviolet

D
The spectral series formula can be derived from the Rutherford model of the hydrogen atom

Solution

Paschen series lies in the infrared region.
We get paschen series of the hydrogen atom. It is obtained in the infrared region. This formula gives a wavelength of lines in the paschen series of the hydrogen spectrum.
Question 9
1 / -0

The fact that photons carry energy was established by

A
Doppler's effect

B
Compton's effect

C
Bohr's theory

D
Diffraction of light

Solution
Question 10
1 / -0

The splitting of line into groups under the effect of electric or magnetic field is called

A
Zeeman's effect

B
Bohr's effect

C
Heisenberg's effect

D
Magnetic effect

Solution

User

Question Analysis

Correct -
Wrong -
Skipped -

My Perfomance

Score
-
out of -
Rank
-
out of -

Re-Attempt Weekly Quiz Competition

Atoms Test - 63

Result

Atoms Test - 63

Score

-

Rank

-

SHARING IS CARING

Question 1

$$1$$

$$2$$

$$3$$

$$4$$

Solution

Question 2

Balmer series

Pfund series

Paschen series

Lyman series

Solution

Question 3

$$+\dfrac{e^{2}}{8\pi\varepsilon_{0}r}$$ and $$-\dfrac{e^{2}}{4\pi\varepsilon_{0}r}$$

$$+\dfrac{8\pi\varepsilon_{0}r}{r}$$ and $$-\dfrac{4\pi\varepsilon_{0}r}{r}$$

$$-\dfrac{e^{2}}{8\pi\varepsilon_{0}r}$$ and $$-\dfrac{e^{2}}{4\pi\varepsilon_{0}r}$$

$$+\dfrac{e^{2}}{8\pi\varepsilon_{0}r}$$ and $$+\dfrac{e^{2}}{4\pi\varepsilon_{0}r}$$

Solution

Question 4

Conservation of linear momentum

Conservation of angular momentum

Conservation of quantum frequency

Conservation of energy

Solution

Question 5

$$n^{-1}$$

$$n$$

$$n^{-2}$$

$$n^{2}$$

Solution

Question 6

Nucleous

Photon

Proton

Atom

Solution

Question 7

$$Z^{2}n^{2}$$

$$\dfrac{Z^{2}}{n^{2}}$$

$$\dfrac{Z^{2}}{n}$$

$$\dfrac{n^{2}}{Z}$$

Solution

Question 8

Lyman series is a continuous spectrum

Paschen series is a line spectrum in the infrared

Balmer series is a line spectrum in the ultraviolet

The spectral series formula can be derived from the Rutherford model of the hydrogen atom

Solution

Question 9

Doppler's effect

Compton's effect

Bohr's theory

Diffraction of light

Solution

Question 10

Zeeman's effect

Bohr's effect

Heisenberg's effect

Magnetic effect

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

Results Announcement on: coming Monday

Stay Tuned: We’ll update you on your result via email or SMS.