### Structure of Atom - Solutions

CBSE Class 11 Chemistry

NCERT Solutions
Chapter 2
STRUCTURE OF THE ATOM

1. (i) Calculate the number of electrons which will together weight one gram.

(ii) Calculate the mass and charge of one mole of electrons.

Ans. (i) Mass of one electron =kg

Number of electrons that weigh kg = 1

Number of electrons that will weigh 1 g =

(ii) Mass of one electron =  kg

Mass of one mole of electron =

kg

Charge on one electron  coulomb

Charge on one mole of electron = $1.6022×{10}^{-19}C×6.022×{10}^{23}$

2. (i) Calculate the total number of electrons present in one mole of methane.

(ii) Find (a) the total number and (b) the total mass of neutrons in 7 mg of

(Assume that mass of a neutron ).

(iii) Find (a) the total number and (b) the total mass of protons in 34 mg of  at STP.

Will the answer change if the temperature and pressure are changed?
Ans. (i) Number of electrons present in 1 molecule of methane

Number of electrons present in 1 mole i.e., $6.022×{10}^{23}$ molecules of methane

$=6.022×{10}^{23}×10=6.022×{10}^{24}$

(ii) (a) Number of atoms of in 1 mole = $6.022×{10}^{23}$

Since 1 atom of  contains (14- 6) i.e., 8 neutrons, the number of neutrons in 14 g of  is $6.022×{10}^{23}×8$  Or, 14 g of contains  neutrons.

Number of neutrons in 7 mg

(b) Mass of one neutron =  kg

Mass of total neutrons in 7 g of

kg

(iii) (a) 1 mole of

= 17 g of

molecules of

Total number of protons present in 1 molecule of

= 10

Number of protons in  molecules of

17 g of  contains  protons.

Number of protons in 34 mg of

(b) Mass of one proton

Total mass of protons in 34 mg of

The number of protons, electrons, and neutrons in an atom is independent of temperature and pressure conditions. Hence, the obtained values will remain unchanged if the temperature and pressure is changed.

3. How many neutrons and protons are there in the following nuclei?

,  ,  ,  ,

Ans.

Atomic mass = 13

Atomic number = Number of protons = 6

Number of neutrons = (Atomic mass) - (Atomic number)

= 13 - 6 = 7

:

Atomic mass = 16

Atomic number = 8

Number of protons = 8

Number of neutrons = (Atomic mass)-(Atomic number)

= 16 -8 = 8

:

Atomic mass = 24

Atomic number = Number of protons = 12

Number of neutrons = (Atomic mass)-(Atomic number)

= 24-12 = 12

:

Atomic mass = 56

Atomic number = Number of protons = 26

Number of neutrons = (Atomic mass) -(Atomic number)

= 56-26 = 30

:

Atomic mass = 88

Atomic number = Number of protons = 38

Number of neutrons = (Atomic mass)-(Atomic number)

= 88 -38 = 50

4. Write the complete symbol for the atom with the given atomic number (Z) andAtomic mass (A)

(i) Z = 17, A = 35

(ii) Z = 92, A = 233

(iii) Z = 4, A = 9

Ans. (i)

(ii)

(iii)

5. Yellow light emitted from a sodium lamp has a wavelength ( ) of 580 nm. Calculate the frequency ( ) and wave number ( ) of the yellow light.

Ans. From the expression,

We get,

…….. (i)

Where,

= frequency of yellow light

c = velocity of light in vacuum

= wavelength of yellow light = 580 nm

Substituting the values in expression (i):

Thus, frequency of yellow light emitted from the sodium lamp

Wave number of yellow light,

6. Find energy of each of the photons which

(i) correspond to light of frequency  Hz.

(ii) have wavelength of 0.50 .

Ans. (i) Energy (E) of a photon is given by the expression,

E =

Where,

h = Planck's constant

v= frequency of light

Substituting the values in the given expression of E:

E =  (ii) Energy (E) of a photon having wavelength is given by the expression,

h = Planck's constant

c = velocity of light in vacuum

Substituting the values in the given expression of E:

7. Calculate the wavelength, frequency and wave number of a light wave whose period is .

Ans. Frequency (v) of light

Wavelength ( ) of light

Where,

c = velocity of light in vacuum

Substituting the value in the given expression of :

Wave number  of light

8. What is the number of photons of light with a wavelength of 4000 pm that provide 1 J of energy?

Ans. Energy (E) of a photon = hv

Energy  of ‘n' photons = nhv

Where,

= wavelength of light = 4000 pm =

c = velocity of light in vacuum =

h = Planck's constant =

Substituting the values in the given expression of n:

Hence, the number of photons with a wavelength of 4000 pm and energy of 1 J are .

9. A photon of wavelength  strikes on metal surface, the work function of the metal being 2.13 eV. Calculate (i) the energy of the photon (eV), (ii) the kinetic energy of the emission, and (iii) the velocity of the photoelectron (1 eV= ).

Ans. (i) Energy (E) of a photon = hv

Where,

h = Planck's constant =

c = velocity of light in vacuum

= wavelength of photon

Substituting the values in the given expression of E:

Hence, the energy of the photon is .

(ii) The kinetic energy of emission is given by

Hence, the kinetic energy of emission is 0.97 eV.

(iii) The velocity of a photoelectron (v) can be calculated by the expression,

Where, is the kinetic energy of emission in Joules and ‘m' is the mass of the photoelectron. Substituting the values in the given expression of v:

Hence, the velocity of the photoelectron is .

10. Electromagnetic radiation of wavelength 242 nm is just sufficient to ionise the sodium atom. Calculate the ionisation energy of sodium in kJ mol-1.

Ans. Energy of sodium (E)

11. A 25 watt bulb emits monochromatic yellow light of wavelength of 0.57 m. Calculate the rate of emission of quanta per second.

Ans. Power of bulb, P= 25 Watt =

Energy of one photon, E= hv

Substituting the values in the given expression of E:

Rate of emission of quanta per second

12. Electrons are emitted with zero velocity from a metal surface when it is exposed to radiation of wavelength 6800  . Calculate threshold frequency ( ) and work function ) of the metal.

Ans. Threshold wavelength of radian  =

Threshold frequency  of the metal

Thus, the threshold frequency of the metal is .

Hence, work function (W0) of the metal

13. What is the wavelength of light emitted when the electron in a hydrogen atom undergoes transition from an energy level with n = 4 to an energy level with n =2?

Ans. The transition will give rise to a spectral line of the Balmer series. The energy involved in the transition is given by the relation,

Substituting the values in the given expression of E:

Wavelength of light emitted

Substituting the values in the given expression of :

14. How much energy is required to ionise a H atom if the electron occupies n = 5 orbit? Compare your answer with the ionization enthalpy of H atom (energy required to remove the electron from n =1 orbit).

Ans. The expression of energy is given by,

Where,

Z = atomic number of the atom

n= principal quantum number

For ionization from to  ,

Hence, the energy required for ionization from n= 5 to n =  is .

Energy required for to n =  ,

On comparing

Hence, less energy is required to ionize an electron in the 5thorbital of hydrogen atom as compared to that in the ground state.

15. What is the maximum number of emission lines when the excited electron of an H atom in n = 6 drops to the ground state?

Ans. When the excited electron of an H atom in n= 6 drops to the ground state, the following transitions are possible:

Hence, a total number of (5 + 4 + 3 + 2 + 1) 15 lines will be obtained in the emission spectrum.

The number of spectral lines produced when an electron in the  level drops down to the ground state is given by .
Given,
n= 6
Number of spectral lines  = 15

16. (i) The energy associated with the first orbit in the hydrogen atom is  atom-1. What is the energy associated with the fifth orbit?

(ii) Calculate the radius of Bohr's fifth orbit for hydrogen atom.
Ans. (i) Energy associated with the fifth orbit of hydrogen atom is calculated as:

(ii) Radius of Bohr's  orbit for hydrogen atom is given by,

For,
n = 5

17. Calculate the wave number for the longest wavelength transition in the Balmer series of atomic hydrogen.
Ans. For the Balmer series,  = 2. Thus, the expression of wave number is given by,

Wave number is inversely proportional to wavelength of transition. Hence, for the longest wavelength transition, has to be the smallest.

For  to be minimum should be minimum. For the Balmer series, a transition from ni = 2 to nf = 3 is allowed. Hence, taking  = 3, we get: