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Standard VIII

Chemistry

Cathode Ray Experiment

The mass of a...

Question

The mass of an electron can be expressed as (A) 0.512 MeV (B) 8.19 × 10–14J/c2 (C) 9.1 × 10–31 kg (D) 0.00055 amu

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Q. The atomic mass of

_{8} O^{16}

15.9949 a m u

. Mass of one neutron and one proton is

2.016490 a m u

and the mass of an electron is

0.00055 a m u

. The binding energy per nucleon of oxygen atom is

The masses of neutron and proton are 1.0087 amu and 1.0073 amu respectively. If the neutrons and protons combine to form a helium nucleus (alpha particles) of mass 4.0015 amu, the binding energy of the helium nucleus will be [1 amu= 931 MeV]

Q. Calculate the mass defect and binding energy per nucleon of

_{8}^{16} O

which has a mass

15.99491 a m u

.
Mass of neutron

= 1.008655 a m u

Mass of proton

= 1.007277 a m u

Mass of electron

= 0.0005486 a m u

1 a m u = 931.5 M e V

Q. In an atom, an electron is moving with a speed of

600 m / s

with an accuracy of

0.005 %

, certainity with which the position of the electron can be located is:

(h = 6.6 \times 10^{- 34}

m^{2} s^{- 1}

, mass of electron

e_{m} = 9.1 \times 10^{- 31}

kg)

Q. Calculate the nuclear binding energy per nucleon of

_{6}^{12} C

from the following data:
Given mass of proton

(m_{p}) = 1.00758

amu
mass of neutron

(m_{n}) = 1.00893

amu
mass of electron

(m_{e}) = 0.00055

amu
Isotopic mass of

^{12} C = 12.00000

amu. Take

1 a m u = 931

MeV
Plan from the mass defect

△ m

(amu), binding energy

= △ m \times 931

MeV
binding energy per nucleon

= \frac{△ m \times 931}{n u m b e r o f n u c l e o n s}

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