Shortcut Methods
JEE Main
| Numerical | Shortcut Method |
|---|---|
| An electron is accelerated through a potential difference of 100 V. Calculate its de Broglie wavelength. | De Broglie equation: $$\lambda =\frac{h}{\sqrt{2meV}}$$ |
| A beam of monochromatic light with a wavelength of 600 nm strikes a metal surface. Calculate the maximum kinetic energy of the emitted electrons. | Einstein’s photoelectric effect equation: $$KE_{max} = h\nu - \phi$$ |
| A hydrogen atom is in the n = 2 energy level. Calculate the wavelength of the photon emitted when the atom transitions to the n = 1 energy level. | Rydberg formula: $$\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$ |
| A nucleus with a mass number of 238 undergoes alpha decay. Calculate the energy released in the process. | Alpha decay energy formula: $$E=-[\left(M_p+M_\alpha\right)c^2-M_d]c^2$$ |
| A neutron is captured by a nucleus with a mass number of 239. Calculate the number of neutrons and protons in the resulting nucleus. | The number of protons remains the same while the number of neutrons increases by 1 in the resulting nucleus. |
CBSE Board Exams
| Numerical | Shortcut Method |
|---|---|
| An electron has a de Broglie wavelength of 0.1 nm. Calculate its kinetic energy | De Broglie equation: $$\lambda =\frac{h}{\sqrt{2meK}}$$ |
| A beam of monochromatic light with a frequency of 5 x 10¹⁴ Hz strikes a metal surface. Calculate the maximum kinetic energy of the emitted electrons | Einstein’s photoelectric effect equation: $$KE_{max} = h\nu -\phi$$ |
| A hydrogen atom is in the n = 3 energy level. Calculate the wavelength of the photon emitted when the atom transitions to the n = 2 energy level | Rydberg formula: $$\frac{1}{\lambda}=R_H\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$$ |
| A nucleus with a mass number of 235 undergoes fission. Calculate the energy released in the process | Nuclear Fission formula: $$E=[(M_A)c^2-(M_C+M_D)c^2)]c^2 $$ |
| A proton is accelerated through a potential difference of 1000 V. Calculate its de Broglie wavelength | De Broglie equation: $$\lambda =\frac{h}{\sqrt{2m_pc^2(\Delta V)}}$$ |
