Atoms Test

Result

Atoms Test - 9

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Weekly Quiz Competition

Question 1
1 / -0

The penetrating power of X-rays increases with the increase in its:

A
both velocity and wavelength

B
velocity

C
frequency

D
wavelength

Solution

both velocity and wavelength
Question 2
1 / -0

The ionisation potential of hydrogen is

A
-10.36 eV

B
13.6 eV

C
14.24 eV

D
8.24 eV

Solution

Ionization potential energy is defined as the minimum energy required in eV to excite a neutral atom to its ionized state i.e it is the minimum energy required in eV to excite an electron from n=1 to infinity. lonization energy is 0-(-13.6 e V)=13.6 eV
Hence, lonization potential is 13.6 eV.
Question 3
1 / -0

The total energy of an electron in the first excited state of hydrogen atom is about - 3.4 eV. Its kinetic energy in this state is:

A
3.4 eV

B
-6.8 eV

C
-3.4 eV

D
6.8 eV

Solution

The kinetic energy of the electron in a orbit is numerically equal to its total energy.
Question 4
1 / -0

When the hydrogen atom is in first excited level, its radius is

A
four times

B
half

C
same

D
twice

Solution

For the first excited state, n = 2
Since r∝ n², the radius of the orbit is four times .
Question 5
1 / -0

Of the following pairs of species, which one will have the same electronic configuration for both members?

A
Li⁺ and Na⁺

B
He and Ne⁺

C
C and N⁺

D
H and Li

Solution

Of all, C and N⁺ have got the same electronic configuration. There are 2 electrons in first orbit and 4 in the second orbit.
Question 6
1 / -0

The Bohr model of atoms

A
uses Einstein's photoelectric equation

B
predicts continuous emission spectra for atoms

C
predicts the same emission spectra for all types of atoms

D
assumes that the angular momentum of electrons is quantised

Solution

assumes that the angular momentum of electrons is quantised

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Re-Attempt Weekly Quiz Competition

Atoms Test - 9

Result

Atoms Test - 9

Score

-

Rank

-

SHARING IS CARING

Question 1

both velocity and wavelength

velocity

frequency

wavelength

Solution

Question 2

-10.36 eV

13.6 eV

14.24 eV

8.24 eV

Solution

Question 3

3.4 eV

-6.8 eV

-3.4 eV

6.8 eV

Solution

Question 4

four times

half

same

twice

Solution

Question 5

Li+ and Na+

He and Ne+

C and N+

H and Li

Solution

Question 6

uses Einstein's photoelectric equation

predicts continuous emission spectra for atoms

predicts the same emission spectra for all types of atoms

assumes that the angular momentum of electrons is quantised

Solution

User

Question Analysis

My Perfomance

Score

-

Rank

-

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Li⁺ and Na⁺

He and Ne⁺

C and N⁺