The total number of nodes for a given principal quantum number $n$, and azimuthal quantum number $\e...

The total number of nodes for a given principal quantum number $ n $ and azimuthal quantum number $ \ell $ is calculated by adding both angular and radial nodes.Angular Nodes: The number of angular nodes is equal to the azimuthal quantum number $ \ell $.Radial Nodes: The number of radial nodes is given by $ n - \ell - 1 $.Thus, the total number of nodes is the sum of angular and radial nodes, which can be expressed as:$ \text{Total Nodes} = \ell + (n - \ell - 1) $This calculation provides the complete count of nodes in an atomic orbital based on the quantum numbers.

$$ \text { How many radial nodes does } \mathrm{Ca}^{+} \text {have in its } 4 \text { s orbital? } ...

To determine the number of radial nodes in the 4s orbital of $\mathrm{Ca}^+$, we use the formula:$ \text{Number of radial nodes} = n - l - 1 $Where:$ n $ is the principal quantum number.$ l $ is the azimuthal quantum number, which is 0 for an s orbital.For the 4s orbital:$ n = 4 $$ l = 0 $Substitute these values into the formula:$ \text{Number of radial nodes} = 4 - 0 - 1 = 3 $Thus, the 4s orbital of $\mathrm{Ca}^+$ has 3 radial nodes.

Atomic Structure - IAT-IISER Chemistry PYQs

Q: Which of the following expressions represents the hydrogen atom wave function $\psi(r)$ shown in the...

On observing the graph, you can observe that $\psi(r)$ has one root.These following equations have only exponential functions of distance from radius and will never be zero as $e$ is a positive number.$$ \frac{1}{\sqrt{\pi}}\left(\frac{1}{a_0}\right)^{3 / 2} e^{-r / 2 a_0} \text { and } \frac{1}{\sqrt{6 \pi}}\left(\frac{1}{a_0}\right)^{3 / 2}\left(\frac{r}{a_0}\right) e^{-r / 2 a_0} $$For the remaining equations, we only need to consider$$ 2-\frac{r}{a_0} \text { and } 3-\frac{2 r}{a_0}+\frac{2 r^2}{9 a_0^2} $$As the rest of the equations are either constants or an exponential function of $r$. Let $\frac{r}{a_0}=x$,$$ 3-\frac{2 r}{a_0}+\frac{2 r^2}{9 a_0^2}=3-2 x+\frac{2 x^2}{9} $$Here, $d=b^2-4 a c=16-4(2 / 9) 3=+v e$,This means it will have two roots whereas the shown graph clearly has only one. Hence,$$ \psi(r)=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{\frac{3}{2}}\left(2-\frac{r}{a_0}\right) e^{-\frac{r}{2 a_0}} $$is the best answer.

Q: The minimum energy needed to remove an electron from a metal corresponds to a wavelength of 500 nm ....

The energy $ E $ of a photon is given by the formula:$ E = \frac{hc}{\lambda} $Where:$ h = 6.6 \times 10^{-34} \, \text{J⋅s} $ (Planck's constant)$ c = 3 \times 10^8 \, \text{m/s} $ (speed of light)$ \lambda = 300 \, \text{nm} = 300 \times 10^{-9} \, \text{m} $Plugging in these values:$ E = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{300 \times 10^{-9}} = 6.6 \times 10^{-18} \, \text{J} $Step 2: Determine the Work Function of the MetalThe work function $ \phi $ is the minimum energy needed to remove an electron from the metal surface, calculated using the wavelength of 500 nm:$ \phi = \frac{hc}{\lambda'} $Where $ \lambda' = 500 \, \text{nm} = 500 \times 10^{-9} \, \text{m} $.$ \phi = \frac{6.6 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} = 3.96 \times 10^{-18} \, \text{J} \approx 4 \times 10^{-18} \, \text{J} $Step 3: Calculate the Kinetic Energy of a Single PhotoelectronThe kinetic energy $ KE $ of an ejected electron is the difference between the photon energy and the work function:$ KE = E - \phi $$ KE = 6.6 \times 10^{-18} - 4 \times 10^{-18} = 2.6 \times 10^{-18} \, \text{J} $Step 4: Find the Number of Photons Emitted per SecondThe power $ P $ of the bulb is the energy emitted per second, given by:$ P = \frac{E_{\text{total}}}{t} $Given:$ P = 100 \, \text{W} = 100 \, \text{J/s} $The number of photons $ N $ emitted per second is:$ N = \frac{100}{6.6 \times 10^{-18}} \approx 15 \times 10^{18} $Step 5: Calculate the Total Kinetic Energy of All Photoelectrons per SecondThe total kinetic energy is the kinetic energy of one photoelectron multiplied by the number of ejected photoelectrons, which equals the number of incident photons:$ \text{Total KE} = N \times KE $$ \text{Total KE} = 15 \times 10^{18} \times 2.6 \times 10^{-18} = 39 \, \text{J} \approx 40 \, \text{J} $Thus, the total kinetic energy of the photoelectrons ejected per second is approximately 40 J.

Q: What is the time period of revolution of an electron in the fourth Bohr orbit of $\mathrm{He}^{+}$?[...

Given dataBohr radius $ a_0 = 52.9\, \text{pm} = 52.9 \times 10^{-12}\, \text{m} $Electron mass $ m_e = 9.11 \times 10^{-31}\, \text{kg} $Planck constant $ h = 6.626 \times 10^{-34}\, \text{J·s} $$\mathrm{He^{+}}$ ⇒ $ Z = 2 $Orbit number $ n = 4 $Step 1: Bohr model relationsFor a hydrogen‐like atom ($Z$),the radius of the $n^{th}$ orbit is$ r_n = \frac{n^2 a_0}{Z} $and the electron speed in that orbit is$ v_n = \frac{Z e^2}{2 \epsilon_0 h} \cdot \frac{2h}{m_e a_0 n} = \frac{Z e^2}{(4 \pi \epsilon_0)} \frac{1}{n \hbar} $—but numerically, we can also use known speed from Bohr model:For hydrogen,$ v_1 = \alpha c = 2.18 \times 10^6\, \text{m/s} $.For a hydrogenic ion,$ v_n = \frac{Z \, v_1}{n} $where $v_1 = 2.18\times10^6\,\mathrm{m/s}$.Step 2: Time period of revolution$ T_n = \frac{2\pi r_n}{v_n} $Substitute the two expressions:$ r_n = \frac{n^2 a_0}{Z}, \quad v_n = \frac{Z v_1}{n} $$ T_n = \frac{2\pi \, (n^2 a_0 / Z)}{(Z v_1 / n)} = \frac{2\pi a_0 n^3}{Z^2 v_1} $Step 3: Substitute numbersFor $\mathrm{He^+}$, $Z = 2$, $n = 4$.$ T_4 = \frac{2\pi a_0 n^3}{Z^2 v_1} = \frac{2\pi (52.9 \times 10^{-12}) (4)^3}{(2)^2 (2.18 \times 10^6)} $Compute step‐by‐step:$ n^3 = 64, \quad Z^2 = 4 $$ T_4 = \frac{2\pi (52.9 \times 10^{-12}) (64)}{4 (2.18 \times 10^6)} = \frac{2\pi (52.9 \times 10^{-12}) (16)}{2.18 \times 10^6} $$ 52.9 \times 16 = 846.4 $$ 2\pi \times 846.4 \times 10^{-12} = 5318 \times 10^{-12} = 5.318 \times 10^{-9} $$ T_4 = \frac{5.318 \times 10^{-9}}{2.18 \times 10^6} = 2.44 \times 10^{-15}\, \text{s} $✅Final Answer:$ T_4 = 2.4 \, \text{femtoseconds} $✅Correct Option:D) 2.4 femtoseconds

01

PYQ 2020

hard

chemistry ID: iat-iise

The total number of nodes for a given principal quantum number, and azimuthal quantum number, are

1

sum of angular nodes and radial nodes

2

sum of angular nodes and radial nodes

3

sum of angular nodes and radial nodes

4

sum of angular nodes and radial nodes

02

PYQ 2022

medium

chemistry ID: iat-iise

Which of the following expressions represents the hydrogen atom wave function shown in the figure below? (is the distance of the electron from the nucleus and is a constant)

1

2

3

4

03

PYQ 2023

easy

chemistry ID: iat-iise

$ $

1

0

2

1

3

2

4

3

04

PYQ 2024

hard

chemistry ID: iat-iise

The minimum energy needed to remove an electron from a metal corresponds to a wavelength of 500 nm . What is the total kinetic energy of all the photoelectrons ejected per second when the entire radiation from a 100 Watt bulb with a wavelength of 300 nm falls on the surface of the metal? [Planck's constant; speed of light]

1

2

3

40 J

4

80 J

05

PYQ 2025

easy

chemistry ID: iat-iise

What is the time period of revolution of an electron in the fourth Bohr orbit of ?[Bohr radius picometers, mass of an electron, Planck's constant]

1

24 femtoseconds

2

4.8 femtoseconds

3

24 femtoseconds

4

2.4 femtoseconds

Atomic Structure

High-Yield Trend

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5 MCQs

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Official Solution

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Official Solution

Atomic Structure

High-Yield Trend

Questions 5 MCQs

Official Solution

Official Solution

Official Solution

Official Solution

Official Solution

Questions
5 MCQs