Tag: nuclei

Quizzes (7) Questions (160)

Questions Related to nuclei

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

A free nucleus of mass $24$ u emits a gamma photon [when initially at rest]. The energy of the photon is $7$ MeV. The recoil energy of the nucleus in keV is 

  1. $1.1$
  2. $1.2$
  3. $1.0$
  4. $1.3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using conservation of momentum, the recoil energy E_r = (E_gamma)^2 / (2 * M * c^2). With E_gamma = 7 MeV and M = 24 u (approx 24 * 931.5 MeV/c^2), the recoil energy is approximately 1.1 keV.

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

Hydrogen atom will be in its ground state,if its electron is in

  1. any energy level

  2. the lowest energy state

  3. the highest energy state

  4. the intermediate state
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Hydrogen atom has one electron. Depending upon the electron residing in which energy state hydrogen energy varies. So, for hydrogen atom to be in ground state electron should be in lowest energy state.

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

When writing electron configurations,electrons are represented in their lowest possible energy state.Which of the following state is this ?

  1. excited state.

  2. lowest state.

  3. configuration state.

  4. ground state.

  5. unenergetic state.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to Aufbau's principle, electrons are filled in the atomic orbitals in order of lower energy to the higher energy starting from the orbital of lowest possible energy known as the ground state.

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

A beam of ultraviolet radiation having wavelength between $100nm$ and $200nm$ is incident on a sample of atomic hydrogen gas. Assuming that the atoms are in ground state, which wavelengths will have low intensity in the transmitted beam? 

  1. $104nm$
  2. $103nm$
  3. $105nm$
  4. $100nm$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Energy corresponding to wavelength 100nm, $\dfrac{1242eV}{100}=12.42 eV$

Energy corresponding to wavelength 200nm, $\dfrac{1242eV}{200}=6.21 eV$
Energy required from ground state to first excited stage:
$E _2-E _1=13.6-3.4=10.2 eV$
Energy required from ground state to second excited stage:
$E _3-E _1=13.6-13.6-1.5=12.1  eV$

Energy required from ground state to third excited stage:
$E _3-E _1=13.6-0.85=12.75  eV$

At 10.2 eV,
$Wavelength 1=\dfrac{1242}{10.2}=122nm$

At 12.1 eV,
$Wavelength 2=\dfrac{1242}{12.1}=103nm$



Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The ground state energy of the electron in hydrogen atom is equal to :

  1. the ground state energy of the electron in $ { He }^{ + } $
  2. the first excited state energy of the electron in $ { He }^{ + } $
  3. the first excited state energy of the electron in $ { Li }^{ +2 } $
  4. the ground state energy of the electron in $ { Be }^{ +3 } $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The energy levels of hydrogen-like ions are given by E = -13.6 * Z^2 / n^2. For H (Z=1, n=1), E = -13.6 eV. For He+ (Z=2), the first excited state is n=2, so E = -13.6 * (2^2) / (2^2) = -13.6 eV.

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The total energy of an electron in the ground state of hydrogen atom is $-13.6\space eV$. The potential energy of an electron in the ground state of $Li^{2+}$ ion will be

  1. $122.4\space eV$
  2. $-122.4\space eV$
  3. $244.8\space eV$
  4. $-244.8\space eV$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Therefore, for ${ Li }^{ 2+ }$ ion, Total energy is $- 13.6\times 9 = 122.4\ eV$
But, $-K.E= T.E= \dfrac { P.E }{ 2 } $
Therefore, $P.E$ is $-244.8\ eV$

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

The ground state energy of Hydrogen atom is $–13.6eV$. The potential energy of the electron in this state is:

  1. $0eV$
  2. $-27.2eV$
  3. $1eV$
  4. $2eV$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ground state energy of hydrogen atom $=13.6eV$

Potential energy $=2$ energy of electron
                             $=2(-13.6\ eV)$
                             $=-27.2\ eV$

Multiple choice physics nuclei gamma decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

In which of the following systems will the radius of the first orbit (n = 1) be minimum ?

  1. hydrogen atom

  2. deuterium atom

  3. singly ionized helium

  4. doubly ionized lithium.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Radius of the first orbit of an atom  $R _1 = \dfrac{0.529}{Z}$  $A^o$
$\implies \ R _1 \propto\dfrac{1}{Z}$
where $Z$ is the atomic number of Hydrogen-like atom.
Since $Z$ is maximum for doubly ionized lithium, thus radius of first orbit is minimum in doubly ionized lithium.

Multiple choice physics nuclei nuclear force the nuclear force nuclear force and binding energy

The binding energy per nucleon for $\mathrm { U } ^ { 238 }$ about 7.5Mev where as it is about 8.5 Mev for a nucleus having a mass half of Uranium. If $\mathrm { U } ^ { 238 }$ splits into two exact halves the energy released would be

  1. 6.4 Mev

  2. 119 Mev

  3. 238 Mev

  4. 476 Mev
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Energy released = (Total binding energy of products) - (Total binding energy of reactants). Reactants: 238 nucleons * 7.5 MeV = 1785 MeV. Products: 2 * (119 nucleons * 8.5 MeV) = 2023 MeV. Energy released = 2023 - 1785 = 238 MeV.