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Question

The number of neutrons present in the atom of 56Be137

The correct answer is

81

Barium Atom Neutron Count Calculation

To find the number of neutrons in an atom, we need to understand the information provided by the atomic notation $ ^{A}_{Z}X $, where:

  • 'X' is the chemical symbol of the element.
  • 'Z' is the atomic number, which represents the number of protons in the nucleus.
  • 'A' is the mass number, which represents the total number of protons and neutrons in the nucleus.

In this question, we are given the atom of Barium as $ ^{137}_{56}Ba $. Here:

  • The atomic number (Z) is 56. This means a Barium atom has 56 protons.
  • The mass number (A) is 137. This means the total number of protons and neutrons is 137.

Calculating Neutrons

The number of neutrons can be calculated using the formula:

Number of neutrons = Mass number (A) - Atomic number (Z)

Using LaTeX for the calculation:

$$ \text{Number of neutrons} = A - Z $$

Substituting the given values for Barium ($ ^{137}_{56}Ba $):

$$ \text{Number of neutrons} = 137 - 56 $$

$$ \text{Number of neutrons} = 81 $$

Therefore, the atom of Barium-137 ($ ^{137}_{56}Ba $) contains 81 neutrons.

Comparing this result with the given options:

  • Option 1: 56 (Incorrect - represents protons)
  • Option 2: 137 (Incorrect - represents mass number)
  • Option 3: 193 (Incorrect calculation)
  • Option 4: 81 (Correct - represents neutrons)

The correct option is 4.

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Important Questions from Atoms

  1. If $M$ is the mass of water that rises in a capillary tube of radius $r$, then what would be the total mass of water that rises if a capillary tube of radius $r$ and another capillary tube of radius $2r$ are simultaneously placed in water, assuming identical liquid and material properties?

  2. The diameter of an atom is

  3. The ratio of specific charge of a proton and a α-particle is

  4. The ratio of radii of two nuclei having atomic mass numbers 27 and 8 respectively, will be:

  5. A $Be^{3+}$ ion, initially in its second excited state, absorbs a photon of wavelength $601.6\text{ A}$. The radius of the ion in the resulting excited state in terms of Bohr radius $a_0$ will be (Take $hc = 12500\text{ eV-A}$)

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