The energy of an electron in the \( n \)-th Bohr orbit is given by the formula: \[ E_n = \frac{-13.6 \, \text{eV}}{n^2}. \] For the first orbit (\( n = 1 \)): \[ E_1 = \frac{-13.6}{1^2} = -13.6 \, \text{eV}. \] For the third orbit (\( n = 3 \)): \[ E_3 = \frac{-13.6}{3^2} = \frac{-13.6}{9} = -1.51 \, \text{eV}. \] Hence, the energy in the third orbit is \( \frac{1}{9} \) of the energy in the first orbit.
List I (Spectral Lines of Hydrogen for transitions from) | List II (Wavelength (nm)) | ||
A. | n2 = 3 to n1 = 2 | I. | 410.2 |
B. | n2 = 4 to n1 = 2 | II. | 434.1 |
C. | n2 = 5 to n1 = 2 | III. | 656.3 |
D. | n2 = 6 to n1 = 2 | IV. | 486.1 |
The angular momentum of an electron in a stationary state of \(Li^{2+}\) (\(Z=3\)) is \( \frac{3h}{\pi} \). The radius and energy of that stationary state are respectively.
Let \( y = f(x) \) be the solution of the differential equation
\[ \frac{dy}{dx} + 3y \tan^2 x + 3y = \sec^2 x \]
such that \( f(0) = \frac{e^3}{3} + 1 \), then \( f\left( \frac{\pi}{4} \right) \) is equal to:
Find the IUPAC name of the compound.
If \( \lim_{x \to 0} \left( \frac{\tan x}{x} \right)^{\frac{1}{x^2}} = p \), then \( 96 \ln p \) is: 32