Home
Class 12
PHYSICS
The atomic masses of the hydrogen isotop...

The atomic masses of the hydrogen isotopes are
Hydrogen `m_1H^1=1.007825` amu
Deuterium `m_1H^2=2.014102` amu
Tritium `m_1H^3=3.016049` amu
The energy released in the reaction,
`_1H^2+_1H^2rarr_1H^3+_1H^1` is nearly

A

(a) 1MeV

B

(b) 2MeV

C

(c) 4MeV

D

(d) 8 MeV

Text Solution

Verified by Experts

The correct Answer is:
C
Energy released `=(Deltam)(931.48)MeV`
`=[2xx2.01102-3.0160-1.007825]xx931.5`
`=4.03MeV=4MeV`
Promotional Banner

Topper's Solved these Questions

  • MODERN PHYSICS - 2

    DC PANDEY|Exercise Level 2 Subjective|10 Videos

  • MODERN PHYSICS - 2

    DC PANDEY|Exercise Level 2 More Than One Correct|6 Videos

  • MODERN PHYSICS - 1

    DC PANDEY|Exercise Level 2 Subjective|23 Videos

  • NUCLEI

    DC PANDEY|Exercise C MADICAL ENTRANCES GALLERY|46 Videos

Similar Questions

Explore conceptually related problems

The atomic masses of the hydrogen isotopes are Hydrogen m_1H^1=1.007825 amu Deuterium m_1H^2=2.014102 amu Tritium m_1H^3=3.016049 amu The number of fusion reactions required to generate 1kWh is nearly

The atomic masses of the hydrogen isotopes are Hydrogen m_1H^1=1.007825 amu Deuterium m_1H^2=2.014102 amu Tritium m_1H^3=3.016049 amu The mass of deuterium, _1H^2 that would be needed to generate 1 kWh

The correct order of wavelength of Hydrogen (._(1)H^(1)) , Deuterium (._(1)H^(2)) and Tritium (._(1)H^(3)) moving with same kinetic energy is

The atomic mass of Li,He, and proton are 7.01823 amu, 4.00387 amu, and 1.00715 amu, respectively. Calculate the energy evolved in the reaction. ._(3)Li^(7) rarr ._(1)P^(1) rarr 2 ._(2)He^(4) + Delta E Given 1 amu = 931 MeV .

Calculate the binding energy of a deutron. Given that mass of proton = 1.007825 amu mass of neutron = 1.008665 amu mass of deutron = 2.014103 amu

Calculate the binding energy for ._(1)H^(2) atom. The mass of ._(1)H^(2) atom is 2.014102 amu where 1n and 1p have their weights 2.016490 amu. Neglect mass of electron.

The atomic masses of deuteron, helium, neutron are 2.014 amu, 3.017 amu and 1.008 amu respectively. On fusion of 0.5 kg of deuterium, ""_(1)H^(2) + ""_(1)H^(2) to ""_(2)He^(3) + ""_(0)n^(1) , the total energy released is

A star initially has 10^40 deuterons. It produces energy via the processes _1^2H+_1^2Hrarr_1^3H+p and _1^2H+_1^3Hrarr_2^4He+n . Where the masses of the nuclei are m( ^2H)=2.014 amu, m(p)=1.007 amu, m(n)=1.008 amu and m( ^4He)=4.001 amu. If the average power radiated by the star is 10^16 W , the deuteron supply of the star is exhausted in a time of the order of

The nuclear reaction .^2H+.^2H rarr .^4 He (mass of deuteron = 2.0141 a.m.u and mass of He = 4.0024 a.m.u ) is

Knowledge Check

  • The atomic masses of the hydrogen isotopes are Hydrogen m_1H^1=1.007825 amu Deuterium m_1H^2=2.014102 amu Tritium m_1H^3=3.016049 amu The number of fusion reactions required to generate 1kWh is nearly

    A
    (a) `10^8`
    B
    (b) `10^18`
    C
    (c) `10^28`
    D
    (d) `10^38`

  • The atomic masses of the hydrogen isotopes are Hydrogen m_1H^1=1.007825 amu Deuterium m_1H^2=2.014102 amu Tritium m_1H^3=3.016049 amu The mass of deuterium, _1H^2 that would be needed to generate 1 kWh

    A
    (a) `3.7kg`
    B
    (b) `3.7g`
    C
    (c) `3.7xx10^-5kg`
    D
    (d) `3.7xx10^-8kg`

  • The correct order of wavelength of Hydrogen (._(1)H^(1)) , Deuterium (._(1)H^(2)) and Tritium (._(1)H^(3)) moving with same kinetic energy is

    A
    `lambda_(H) gt lambda_(D) gt lambda_(r)`
    B
    `lambda_(H)=lambda_(D)=lambda_(r)`
    C
    `lambda_(H) lt lambda_(D) lt lambda_(r)`
    D
    `lambda_(H) lt lambda_(D) gt lambda_(r)`