Atomic Structure

The total number of nodes for a given principal quantum number and azimuthal quantum number is calculated by adding both angular and radial nodes.Angular Nodes: The number of angular nodes is equal to the azimuthal quantum number .Radial Nodes: The number of radial nodes is given by .Thus, the total number of nodes is the sum of angular and radial nodes, which can be expressed as: This calculation provides the complete count of nodes in an atomic orbital based on the quantum numbers.

On observing the graph, you can observe that has one root.These following equations have only exponential functions of distance from radius and will never be zero as is a positive number.$ r \frac{r}{a_0}=x d=b^2-4 a c=16-4(2 / 9) 3=+v e $is the best answer.

To determine the number of radial nodes in the 4s orbital of , we use the formula: Where: is the principal quantum number. is the azimuthal quantum number, which is 0 for an s orbital.For the 4s orbital: Substitute these values into the formula: Thus, the 4s orbital of has 3 radial nodes.

The energy of a photon is given by the formula: Where: (Planck's constant) (speed of light) Plugging in these values: Step 2: Determine the Work Function of the MetalThe work function is the minimum energy needed to remove an electron from the metal surface, calculated using the wavelength of 500 nm: Where . Step 3: Calculate the Kinetic Energy of a Single PhotoelectronThe kinetic energy of an ejected electron is the difference between the photon energy and the work function: Step 4: Find the Number of Photons Emitted per SecondThe power of the bulb is the energy emitted per second, given by: Given: The number of photons emitted per second is: Step 5: Calculate the Total Kinetic Energy of All Photoelectrons per SecondThe total kinetic energy is the kinetic energy of one photoelectron multiplied by the number of ejected photoelectrons, which equals the number of incident photons: Thus, the total kinetic energy of the photoelectrons ejected per second is approximately 40 J.

Given dataBohr radius Electron mass Planck constant ⇒ Orbit number Step 1: Bohr model relationsFor a hydrogen‐like atom ( ),the radius of the orbit is and the electron speed in that orbit is —but numerically, we can also use known speed from Bohr model:For hydrogen, .For a hydrogenic ion, where .Step 2: Time period of revolution Substitute the two expressions: Step 3: Substitute numbersFor , , . Compute step‐by‐step: ✅Final Answer: ✅Correct Option:D) 2.4 femtoseconds