Home
Class 12
CHEMISTRY
The mass of a proton is 1836 times more ...

The mass of a proton is `1836` times more than the mass of an electron. It a sub-atomic of mass `(m) 207` times the mass of electron is captured by the nucleus, then the first ionization potential of `H`:

A

decreases

B

increases

C

remains same

D

may be decreases ot increase

Text Solution

AI Generated Solution

The correct Answer is:
To solve the problem, we need to determine how the first ionization potential of hydrogen changes when a subatomic particle of mass \( m \) (207 times the mass of an electron) is captured by the nucleus. ### Step-by-Step Solution: 1. **Understand the Masses Involved**: - The mass of a proton (\( m_p \)) is given as \( 1836 \) times the mass of an electron (\( m_e \)). - The mass of the subatomic particle (\( m \)) is given as \( 207 \) times the mass of an electron. 2. **Express the Masses**: - Let \( m_e \) be the mass of the electron. - Then, \( m_p = 1836 \times m_e \). - The mass of the subatomic particle is \( m = 207 \times m_e \). 3. **Calculate the Reduced Mass**: - The reduced mass (\( \mu \)) of the system consisting of the proton and the captured subatomic particle can be calculated using the formula: \[ \mu = \frac{m_p \cdot m}{m_p + m} \] - Substituting the values: \[ \mu = \frac{(1836 \times m_e) \cdot (207 \times m_e)}{(1836 \times m_e) + (207 \times m_e)} \] 4. **Simplify the Expression**: - Factor out \( m_e \) from the numerator and denominator: \[ \mu = \frac{1836 \cdot 207 \cdot m_e^2}{(1836 + 207) \cdot m_e} \] - This simplifies to: \[ \mu = \frac{1836 \cdot 207 \cdot m_e}{1836 + 207} \] 5. **Calculate the Total Mass**: - Calculate \( 1836 + 207 = 2043 \). - Now, substituting this back into the equation gives: \[ \mu = \frac{1836 \cdot 207 \cdot m_e}{2043} \] 6. **Determine the Ionization Potential**: - The ionization potential (\( I \)) is directly proportional to the reduced mass (\( \mu \)): \[ I \propto \mu \] - Since the reduced mass has increased due to the addition of the subatomic particle, the ionization potential of hydrogen will also increase. 7. **Conclusion**: - Therefore, the first ionization potential of hydrogen will increase. ### Final Answer: The first ionization potential of hydrogen will increase.

To solve the problem, we need to determine how the first ionization potential of hydrogen changes when a subatomic particle of mass \( m \) (207 times the mass of an electron) is captured by the nucleus. ### Step-by-Step Solution: 1. **Understand the Masses Involved**: - The mass of a proton (\( m_p \)) is given as \( 1836 \) times the mass of an electron (\( m_e \)). - The mass of the subatomic particle (\( m \)) is given as \( 207 \) times the mass of an electron. ...
Promotional Banner

Topper's Solved these Questions

  • NUCLEAR CHEMISTRY

    RESONANCE ENGLISH|Exercise PART-II SECTION-1|21 Videos
  • NUCLEAR CHEMISTRY

    RESONANCE ENGLISH|Exercise PART-III|5 Videos
  • NUCLEAR CHEMISTRY

    RESONANCE ENGLISH|Exercise PART-II|26 Videos
  • NITROGEN CONTAINING COMPOUNDS

    RESONANCE ENGLISH|Exercise ORGANIC CHEMISTRY(Nitrogen containing Compounds)|30 Videos
  • P BLOCK ELEMENTS

    RESONANCE ENGLISH|Exercise PART -II|23 Videos
RESONANCE ENGLISH-NUCLEAR CHEMISTRY-ADVANCED LEVEL PROBLEMS
  1. The increasing order (lowest first) for the values of e//m (charge/mas...

    Text Solution

    |

  2. An electron in an atom jumps in such a way that its kinetic energy cha...

    Text Solution

    |

  3. What atomic number of an element "X" would have to become so that the ...

    Text Solution

    |

  4. Select the incorrect graph for velocity of e^(-) in an orbit vs. Z, 1/...

    Text Solution

    |

  5. Which of the following is discreted in Bohr's theory?

    Text Solution

    |

  6. The mass of a proton is 1836 times more than the mass of an electron. ...

    Text Solution

    |

  7. In any subshell, the maimum number of electrons having same value of s...

    Text Solution

    |

  8. Which quantum number defines the orientation of orbital in the space a...

    Text Solution

    |

  9. For similar orbitals having different values of n:

    Text Solution

    |

  10. Maximum number of nodes are present in :

    Text Solution

    |

  11. The correct set of quantum numbers for the unpaired electron of chlori...

    Text Solution

    |

  12. Which of the following has the maximum number of unpaired electrons?

    Text Solution

    |

  13. Calculate the angular frequency of an electron occupying the second ...

    Text Solution

    |

  14. An excited state of H atom emits a photon of wavelength lamda and retu...

    Text Solution

    |

  15. Light of wavelength lambda shines on a metal surface with initail X an...

    Text Solution

    |

  16. Neutron scattering experiments have shown that the radius of the nucle...

    Text Solution

    |

  17. The nucleus of an atom is located at x=y=z=0. If the probability of fi...

    Text Solution

    |

  18. The energy of a I,II and III energy levels of a certain atom are E, (4...

    Text Solution

    |

  19. A compound of vanadium has magnetic moment of 1.73 BM. Work out the el...

    Text Solution

    |

  20. Calculate the minimum and maximum number of electrons which may have m...

    Text Solution

    |