Home
Class 12
PHYSICS
When an aluminium nucleus (.13Al^27) is ...

When an aluminium nucleus `(._13Al^27)` is hit by a proton a new element is formed with the emission of `alpha`-particle. (ii)Write the complete equation of reaction , (ii)Identify the new element and (iii)Determine the number of neutrons and protons in the nucleus .

Text Solution

Verified by Experts

Proton :`._1H^1`, `alpha`-particle : `._2He^4`
`therefore` Let the new element =`._ZX^A` [ where A = mass number , Z= atomic number ]
From the laws of conservation of mass and atomic number,
27+1 = A+4 and 13+1 = Z+2
or, A=24 and Z = 12
i.The complete equation of the reaction : `._13Al^27+._1^1H to ._12Mg^24 + ._2He^4`
ii.As Z =12 , the element is magnesium (Mg).
So, new element `=._12Mg^24`
iii. Now , we know , mass number = proton number + neutron number (x)
or 24 = 12 + x [ `because` proton number = atomic number ]
or, x=12
Promotional Banner

Topper's Solved these Questions

  • ATOMIC NUCLEUS

    CHHAYA PUBLICATION|Exercise SECTION RELATED QUESTIONS|54 Videos

  • ATOMIC NUCLEUS

    CHHAYA PUBLICATION|Exercise HOTS QUESTION|14 Videos

  • ATOM

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|18 Videos

  • CAPACITANCE AND CAPACITOR

    CHHAYA PUBLICATION|Exercise CBSE SCANNER|15 Videos

Similar Questions

Explore conceptually related problems

When ._4Be^9 is hit by alpha -particles a neutron is emitted resulting in the formation of a new element. Identify the element and write the complete reaction equation.

Explain why a new element is formed by alpha -particle emission but no new element is formed by gamma - ray emission form a radioactive element.

When Be is bombarded with alpha -particles, a grew element viz carbon is formed whereas, when gold is bombarded with alpha -particles, no new elements are formed. Explain.

._92U^238 decays by emitting successively 8 alpha -particles and 8 beta -particles. Determine the mass number and atomic number of the new element and express it in symbol.

When an alpha - particle is emitted from a uranium nucleus (atomic no. 92, mass number 238), a new nucleus is formed. From this nucleus a beta -particle is also emitted. Wha will be the atomic number and mass number of the final nucleus ?

On collision with neutron , ._13Al^27 changes to radiosodium ._11Na^24 and emits a particle. ._11Na^24 , In its turn emits a particle and is transmitted to ._12Mg^24 . Write the two nuclear equations and identify the particles .

Statement I:Some energy is released when a heavy nucleus disintegrate into two nuclei of moderate size. Statement II: The more the mass number of the nucleus , more is the binding energy for each proton or neutron

An element occupies group no.13 and period number 2 is an representative element of that group reacts with carbondioxide and forms an oxide (A). (A) reacts with CuSO_(4) give blue beads (B). Identify the element compound (A) and (B). Write the reaction.

The french physicist Louis de Broglie in 1924 postulated that matter like radiation show a dual behaviour . He proposed the following relationship between the wavelength lambda of a material particle its linear momentum p and planck constant h lamda=h/p =h/(mv) The de broglie relation implies that the wavelength of a particle should decreases as its velocity increases . it also implies that for a given velocity heavier particles should have shorter wavelength than lighter particles. The waves associated with particles in motion are called matter waves or de broglie waves. These waves differ from the electromagnetic waves as they (i) have lower velocities (ii) have no electrical and magnetic fields and (iii) are not emitted by the particle under consideration . The experimental confirmation of the de-broglie relation was obtained when Davisson ans Germer in 1927 observed that a beam of electrons is diffracted by a nickel crystal . as diffraction a characteristics property of waves hence the beam of electron behaves as a wave, as proposed by de-broglie. If proton, electron and alpha -particle are moving with same kinetic energy then the order of de-Broglie's wavelength