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Numerical 1: De Broglie wavelength of an electron The de Broglie wavelength of an electron is given by the formula: $$\lambda =\frac{h}{p}$$ where h is the Planck’s constant and p is the momentum of the electron. The momentum of the electron can be calculated using the formula: $$p=mv$$ where m is the mass of the electron and v is its velocity. In this case, the electron is accelerated through a potential difference of 100 V, so its kinetic energy is: $$K=eV$$ where e is the charge of the electron and V is the potential difference. The kinetic energy of the electron can be used to calculate its momentum: $$p=\sqrt{2mK}$$ Substituting this into the formula for the de Broglie wavelength, we get: $$\lambda =\frac{h}{\sqrt{2meV}}$$ Substituting the values of h, m, e, and V, we get: $$\lambda =\frac{6.626\times {10}^{-34}}{(2\times 9.109\times {10}^{-31}\times 1.602\times {10}^{-19}\times 100})m$$ $$\lambda =1.22nm$$

Numerical 2: Kinetic energy of neutrons The kinetic energy of a neutron can be calculated using the formula: $$K=\frac{p^2}{2m}$$ where p is the momentum of the neutron and m is its mass. The momentum of the neutron can be calculated using the formula: $$p=h/\lambda$$ where h is the Planck’s constant and λ is the wavelength of the neutron beam. In this case, the neutron beam has a wavelength of 1.5 nm, so its momentum is: $$p=h/\lambda =\frac{6.626\times {10}^{-34}}{1.5\times {10}^{-9}}=4.42\times {10}^{-25}$$ Substituting this into the formula for the kinetic energy, we get: $$K=\frac{p^2}{2m}=\frac{{4.42}^2\times {10}^{-50}}{2\times 1.6749\times {10}^{-27}}$$ $$K=6.16\times {10}^{-25}$$ Converting to eV, we get: $$K=6.16\times {10}^{-25}\times 6.24\times {10}^{18}$$ $$K=3.86\times {10}^{-6}eV$$

Numerical 3: De Broglie wavelength of a hydrogen atom The de Broglie wavelength of an object can be calculated using the formula: $$\lambda =\frac{h}{p}$$ where h is the Planck’s constant and p is the momentum of the object. The momentum of the hydrogen atom can be calculated using the formula: $$p=mv$$ where m is the mass of the hydrogen atom and v is its velocity. In this case, the hydrogen atom has a mass of 1.6749×10^−27 kg and a velocity of 1 m/s, so its momentum is: $$p=1.6749\times {10}^{-27}\times 1=1.6749\times {10}^{-27}$$ Substituting this into the formula for the de Broglie wavelength, we get: $$\lambda =\frac{6.626\times {10}^{-34}}{1.6749\times {10}^{-27}}=3.95\times {10}^{-7}$$ Converting to Å, we get: $$\lambda =3.95\times {10}^{-7}\times {10}^{10}$$ $$\lambda =395\text{Å}$$

Numerical 4: Energy of the electron in the first excited state of hydrogen The energy of an electron in a hydrogen atom can be calculated using the formula: $$E=-13.6eV(\frac{1}{n^2}-\frac{1}{m^2})$$ where n is the principal quantum number and m is the angular momentum quantum number. In this case, the electron is in the first excited state, so n = 2 and m = 1. Substituting these values into the formula, we get: $$E=-13.6eV(\frac{1}{2^2}-\frac{1}{1^2})$$ $$E=-13.6(\frac{1}{4}-1)$$ $$E=10.2eV$$

Numerical 5: Wavelength of light emitted when an electron in a hydrogen atom jumps from the second excited state to the first excited state When an electron in a hydrogen atom jumps from the second excited state (n = 3) to the first excited state (n = 2), the wavelength of the light emitted can be calculated using the formula: $$\frac{1}{\lambda }=R(\frac{1}{n_1^2}-\frac{1}{n_2^2})$$ where R is the Rydberg constant and n1 and n2 are the principal quantum numbers of the two states. In this case, n1 = 2 and n2 = 3, so the wavelength of the light emitted is: $$\frac{1}{\lambda }=1.097\times {10}^7(\frac{1}{2^2}-\frac{1}{3^2})$$ $$\frac{1}{\lambda }=1.097\times {10}^7\left(\frac{5}{36}\right)$$ $$\lambda =656nm$$

CBSE Board Exam

Numerical 1: De Broglie wavelength of an electron The de Broglie wavelength of an electron is given by the formula: $$\lambda =\frac{h}{p}$$ where h is the Planck’s constant and p is the momentum of the electron. The momentum of the electron can be calculated using the formula: $$p=mv$$ where m is the mass of the electron and v is its velocity. In this case, the electron has a mass of 9.109×10^-31 kg and a velocity of 1 m/s, so its momentum is: $$p=\sqrt{2mK}$$ Substituting this into the formula for the de Broglie wavelength, we get: $$\lambda =\frac{h}{\sqrt{2meV}}$$ Substituting the values of h, m, e, and V, we get: $$\text{Missing argument for frac}$$ $$\lambda =3.86\times {10}^{-12}m$$

Numerical 2: Kinetic energy of neutrons The kinetic energy of a neutron can be calculated using the formula: $$K=\frac{p^2}{2m}$$ where p is the momentum of the neutron and m is its mass. The momentum of the neutron can be calculated using the formula: $$p=h/\lambda$$ where h is the Planck’s constant and λ is the wavelength of the neutron beam. In this case, the neutron beam has a wavelength of 2 nm, so its momentum is: $$p=h/\lambda =\frac{6.626\times {10}^{-34}}{2\times {10}^{-9}}=3.31\times {10}^{-25}$$ Substituting this into the formula for the kinetic energy, we get: $$K=\frac{p^2}{2m}=\frac{3.31\times {10}^{-25}}{2\times 1.6749\times {10}^{-27}}$$ $$K=1.09\times {10}^{-25}J$$ Converting to eV, we get: $$K=1.09\times {10}^{-25}\times 6.24\times {10}^{18}$$ $$K=6.82\times {10}^{-7}eV$$

Numerical 3: De Broglie wavelength of an atom The de Broglie wavelength of an object can be calculated using the formula: $$\lambda =\frac{h}{p}$$ where h is the Planck’s constant and p is the momentum of the object. The momentum of the atom can be calculated using the formula: $$p=mv$$ where m is the mass of the atom and v is its velocity. In this case, the atom has a mass of 1.6749×10^-27 kg and a velocity of 1 m/s, so its momentum is: $$p=mv=1.6749\times {10}^{-27}$$ Substituting this into the formula for the de Broglie wavelength, we get: $$\lambda = \frac{6.