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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: $$\lambda=\frac{6.626\times10^{−34}}{(2\times9.109 \times10^{−31}\times1.602\times10^{−19}\times100}) m$$ $$\lambda=1.22 nm$$

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: $$p=h/\lambda=\frac{6.626\times10^{−34}}{1.5\times10^{−9}}=4.42\times10^{−25}$$ Substituting this into the formula for the kinetic energy, we get: $$K=\frac{p^2}{2m}=\frac{4.42^2\times10^{−50}}{2 \times 1.6749 \times 10^{−27}}$$ $$K=6.16\times10^{ -25}$$ Converting to eV, we get: $$K=6.16\times10^{ −25}\times 6.24\times10^{18}$$ $$K=3.86\times10^{ -6}eV$$

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: $$p=1.6749 \times 10^{-27}\times1=1.6749 \times 10^{-27}$$ Substituting this into the formula for the de Broglie wavelength, we get: $$\lambda = \frac{6.626\times10^{−34}}{1.6749 \times 10^{-27}}=3.95 \times 10^{-7}$$ Converting to Å, we get: $$\lambda =3.95\times10^{−7}\times 10^{10}$$ $$\lambda = 395 \text {Å}$$

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: $$E = -13.6eV \left(\frac{1}{n^2} - \frac{1}{m^2}\right)$$ 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: $$E = -13.6eV\left(\frac{1}{2^2} - \frac{1}{1^2}\right)$$ $$E = -13.6\left(\frac{1}{4} - 1\right)$$ $$E = 10.2 eV$$

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: $$\frac{1}{\lambda}= R\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)$$ 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: $$\frac{1}{\lambda}= 1.097\times10^7 \left(\frac{1}{2^2} - \frac{1}{3^2}\right)$$ $$\frac{1}{\lambda}= 1.097 \times10^7 \left(\frac{5}{36}\right)$$ $$\lambda= 656 nm$$

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: $$\lambda=\frac{6.626\times10^{−34}{(2\times9.109 \times10^{−31}\times1.602\times10^{−19}\times1})}$$ $$\lambda= 3.86\times10^{-12} m$$

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: $$p=h/\lambda=\frac{6.626\times10^{−34}}{2\times10^{−9}}=3.31 \times 10^{-25}$$ Substituting this into the formula for the kinetic energy, we get: $$K=\frac{p^2}{2m}=\frac{3.31 \times 10^{-25}}{2 \times 1.6749 \times 10^{−27}}$$ $$K=1.09\times10^{ -25}J$$ Converting to eV, we get: $$K=1.09\times10^{ -25}\times 6.24\times10^{18}$$ $$K=6.82 \times 10^{ -7} eV$$

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: $$p=mv= 1.6749\times10^{−27}$$ Substituting this into the formula for the de Broglie wavelength, we get: $$\lambda = \frac{6.