All Important Formula — Class 11 Chemistry Notes
All Important Formula · Class 11 Chemistry · 3 topics.
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Topics covered in All Important Formula
1.Kinetic Theory of Gases Formulas
1. Boyle’s Law
Formula:
PV=constant (at constant temperature)PV = \text{constant (at constant temperature)}
Explanation:Boyle's law states that the pressure (P) of a gas is inversely proportional to its volume (V) if the temperature remains constant. If you decrease the volume, the pressure increases.
Example:If a gas occupies 2 liters at a pressure of 3 atm, what will the volume be if the pressure increases to 6 atm?
P1V1=P2V2P_1V_1 = P_2V_2 3×2=6×V2 ⟹ V2=1 liter3 \times 2 = 6 \times V_2 \implies V_2 = 1 \text{ liter}
2. Charles’ Law
Formula:
VT=constant (at constant pressure)\frac{V}{T} = \text{constant (at constant pressure)}
Explanation:
Charles’ law states that the volume (V) of a gas is directly proportional to its absolute temperature (T) when pressure is kept constant.
Example:If a gas has a volume of 3 liters at 300 K, what will the volume be at 600 K?
V1T1=V2T2\frac{V_1}{T_1} = \frac{V_2}{T_2} 3300=V2600 ⟹ V2=6 liters\frac{3}{300} = \frac{V_2}{600} \implies V_2 = 6 \text{ liters}
3. Ideal Gas Equation
Formula:
PV=nRTPV = nRT
Explanation:
This equation relates pressure (P), volume (V), number of moles (n), universal gas constant (R), and temperature (T) of an ideal gas.
Example:Calculate the pressure of 2 moles of gas at 300 K occupying a volume of 10 liters.
(R = 0.0821 L·atm/mol·K)P=nRTVP = \frac{nRT}{V} P=2×0.0821×30010 ⟹ P=4.926 atmP = \frac{2 \times 0.0821 \times 300}{10} \implies P = 4.926 \text{ atm}
4. Mean Square Velocity
Formula:
C2=C12+C22+C32+…+Cn2nC^2 = \frac{C_1^2 + C_2^2 + C_3^2 + \ldots + C_n^2}{n}
Explanation:
The mean square velocity is the average of the squares of the velocities of gas molecules.
Example:For three molecules with velocities 2 m/s, 3 m/s, and 4 m/s:
C2=22+32+423=4+9+163=293≈9.67 m2/s2C^2 = \frac{2^2 + 3^2 + 4^2}{3} = \frac{4 + 9 + 16}{3} = \frac{29}{3} \approx 9.67 \text{ m}^2/\text{s}^2
5. Pressure of a Gas
Formula:
P=13ρC2P = \frac{1}{3} \rho C^2
Explanation:
This formula relates the pressure (P) of a gas to its density (ρ\rho) and the mean square velocity (C2C^2).
Example:If the density of a gas is 0.5 kg/m³ and the mean square velocity is 300 m²/s:
P=13×0.5×300=50 PaP = \frac{1}{3} \times 0.5 \times 300 = 50 \text{ Pa}
6. Kinetic Energy per Unit Volume
Formula:
KE=32P\text{KE} = \frac{3}{2} P
Explanation:
The kinetic energy per unit volume of a gas is proportional to its pressure.
Example:If the pressure is 100 Pa:
KE=32×100=150 J/m3\text{KE} = \frac{3}{2} \times 100 = 150 \text{ J/m}^3
7. Gas Equation
Formula:
PV=13mNC2PV = \frac{1}{3} mNC^2
Explanation:
This equation relates the pressure (P) and volume (V) of a gas to the number of molecules (N), mass (m), and mean square velocity (C2C^2).
8. Kinetic Energy of a Mole of Gas
Formula:
KE=32RT\text{KE} = \frac{3}{2} RT
Explanation:
The kinetic energy of one mole of gas depends on the temperature (T) and gas constant (R).
Example:At 300 K (R = 8.314 J/mol·K):
KE=32×8.314×300=3741.3 J\text{KE} = \frac{3}{2} \times 8.314 \times 300 = 3741.3 \text{ J}
9. Kinetic Energy of a Molecule
Formula:
KE=32kT\text{KE} = \frac{3}{2} kT
Explanation:
This gives the kinetic energy of a single molecule, where kk is the Boltzmann constant.
10. Mayer’s Formula
Formula:
Cp−Cv=RC_p - C_v = R
Explanation:
This formula shows the relationship between the specific heat at constant pressure (CpC_p) and constant volume (CvC_v) for a gas.
2.Atoms, Molecules, and Nuclei Formulas
1. Radioactive Decay
Formula:
N=N0e−λtN = N_0 e^{-\lambda t}
Explanation:
This formula gives the number of undecayed nuclei (NN) at time tt.
- N0N_0 = Initial number of nuclei
- λ\lambda = Decay constant
- tt = Time
Example:
If N0=1000N_0 = 1000 and λ=0.001\lambda = 0.001 per second, find NN after 1000 seconds:
N=1000×e−0.001×1000=1000×e−1≈368N = 1000 \times e^{-0.001 \times 1000} = 1000 \times e^{-1} \approx 368
2. Half-Life Period
Formula:
T1/2=0.693λT_{1/2} = \frac{0.693}{\lambda}
Explanation:
This formula gives the half-life (T1/2T_{1/2}), which is the time taken for half of the radioactive substance to decay.
Example:If the decay constant λ\lambda is 0.001 per second:
T1/2=0.6930.001=693 secondsT_{1/2} = \frac{0.693}{0.001} = 693 \text{ seconds}
3. Decay Constant Formula
Formula:
λ=2.303tlogN0N\lambda = \frac{2.303}{t} \log \frac{N_0}{N}
Explanation:
This formula calculates the decay constant (λ\lambda) using the initial and remaining number of nuclei over time tt.
Example:If N0=1000N_0 = 1000, N=500N = 500, and t=10t = 10 seconds:
λ=2.30310log1000500≈0.0693 per second\lambda = \frac{2.303}{10} \log \frac{1000}{500} \approx 0.0693 \text{ per second}
4. Second Postulate of Bohr's Theory
Formula:
mvr=nh2πmvr = \frac{nh}{2\pi}
Explanation:
This formula states that the angular momentum of an electron in a stable orbit is quantized.
- mm = Mass of the electron
- vv = Velocity
- rr = Radius of the orbit
- nn = Principal quantum number
- hh = Planck's constant
5. Bohr’s Formula for Wavelength of Emitted Light
Formula:
1λ=R(1n12−1n22)\frac{1}{\lambda} = R \left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)
Explanation:
This formula calculates the wavelength (λ\lambda) of the emitted photon during an electron transition between two energy levels (n1n_1 and n2n_2).
- RR = Rydberg constant
Example:
For n1=1n_1 = 1 and n2=2n_2 = 2 (Lyman series):
1λ=R(112−122)=R(1−0.25)=0.75R\frac{1}{\lambda} = R \left(\frac{1}{1^2} - \frac{1}{2^2}\right) = R \left(1 - 0.25\right) = 0.75 R
6. Einstein's Energy-Mass Equation
Formula:
E=mc2E = mc^2
Explanation:
This formula shows the relationship between energy (EE) and mass (mm), where cc is the speed of light.
Example:If m=1m = 1 kg:
E=1×(3×108)2=9×1016 joulesE = 1 \times (3 \times 10^8)^2 = 9 \times 10^{16} \text{ joules}
7. Radius of nthn^{th} Bohr Orbit
Formula:
rn=n2h24π2me2Zr_n = \frac{n^2 h^2}{4\pi^2 m e^2 Z}
Explanation:
This gives the radius of the nthn^{th} orbit in a hydrogen-like atom.
- ZZ = Atomic number
8. De Broglie Wavelength of Electron
Formula:
λ=hp=hmv=h2meV\lambda = \frac{h}{p} = \frac{h}{mv} = \frac{h}{\sqrt{2meV}}
Explanation:
This formula calculates the wavelength (λ\lambda) of a particle with momentum pp.
Example:For an electron with velocity v=106v = 10^6 m/s (h = 6.63×10−346.63 \times 10^{-34} Js, m = 9.1×10−319.1 \times 10^{-31} kg):
λ=6.63×10−349.1×10−31×106≈7.3×10−10 m\lambda = \frac{6.63 \times 10^{-34}}{9.1 \times 10^{-31} \times 10^6} \approx 7.3 \times 10^{-10} \text{ m}
9. Number of Photons
Formula:
n=Pλhcn = \frac{P \lambda}{hc}
Explanation:
This formula calculates the number of photons (nn) in a beam of light with power PP and wavelength λ\lambda.
Example:For P=3P = 3 W, λ=500 nm=500×10−9\lambda = 500 \text{ nm} = 500 \times 10^{-9} m:
n=3×500×10−96.63×10−34×3×108≈7.5×1015
3.Electrons and Photons Formulas
1. Velocity of Electron
Formula:
V=EBV = \frac{E}{B}
Explanation:
This formula gives the velocity (VV) of an electron moving in an electric field (EE) and a magnetic field (BB).
2. Energy of a Photon
Formula:
E=hν=hcλE = h\nu = \frac{hc}{\lambda}
Explanation:
This formula gives the energy (EE) of a photon in terms of Planck's constant (hh), frequency (ν\nu), speed of light (cc), and wavelength (λ\lambda).
Example:If λ=500\lambda = 500 nm,
h=6.63×10−34h = 6.63 \times 10^{-34} Js, c=3×108c = 3 \times 10^8 m/s:E=6.63×10−34×3×108500×10−9≈3.98×10−19 JE = \frac{6.63 \times 10^{-34} \times 3 \times 10^8}{500 \times 10^{-9}} \approx 3.98 \times 10^{-19} \text{ J}
3. Wavelength and Frequency Relationship
Formula:
λ=cν\lambda = \frac{c}{\nu}
Explanation:
This formula relates the wavelength (λ\lambda) of a photon to its frequency (ν\nu) and the speed of light (cc).
4. Einstein’s Photoelectric Equation
Formula:
EP=hν=12mv2+hν0E_P = h\nu = \frac{1}{2} mv^2 + h\nu_0
Explanation:
This equation relates the energy of a photon (hνh\nu) to the kinetic energy of the emitted electron and the work function (hν0h\nu_0).
Example:If hν=5 eVh\nu = 5 \, \text{eV} and hν0=2 eVh\nu_0 = 2 \, \text{eV}:
12mv2=5−2=3 eV\frac{1}{2} mv^2 = 5 - 2 = 3 \, \text{eV}
5. Threshold Frequency
Formula:
ν0=he\nu_0 = \frac{h}{e}
Explanation:
This formula represents the threshold frequency (ν0\nu_0) required to emit an electron during the photoelectric effect.
6. Maximum Kinetic Energy of Electron
Formula:
K.Emax=12mvmax2=h(ν−ν0)K.E_{\text{max}} = \frac{1}{2} mv^2_{\text{max}} = h \left(\nu - \nu_0\right)
Explanation:
This formula gives the maximum kinetic energy of an electron emitted in the photoelectric effect.
Example:If hν=4h\nu = 4 eV and hν0=2h\nu_0 = 2 eV:
K.Emax=4−2=2 eVK.E_{\text{max}} = 4 - 2 = 2 \, \text{eV}
7. Kinetic Energy and Frequency Relationship
Formula:
K.Emax2=hν2−hν02\frac{K.E_{\text{max}}}{2} = \frac{h\nu}{2} - \frac{h\nu_0}{2}
Explanation:
This formula expresses kinetic energy as a function of photon frequency and threshold frequency.
8. Electron Velocity Formula
Formula:
12mv2=KEandv=2KEm\frac{1}{2} mv^2 = \text{KE} \quad \text{and} \quad v = \sqrt{\frac{2 \text{KE}}{m}}
Explanation:
This formula calculates the velocity of an electron when its kinetic energy (KE\text{KE}) is known.
Example:If KE=4×10−19\text{KE} = 4 \times 10^{-19} J and m=9.1×10−31m = 9.1 \times 10^{-31} kg:
v=2×4×10−199.1×10−31≈9.4×105 m/sv = \sqrt{\frac{2 \times 4 \times 10^{-19}}{9.1 \times 10^{-31}}} \approx 9.4 \times 10^5 \text{ m/s}
9. Wavelength of Electron
Formula:
λ=h2meV\lambda = \frac{h}{\sqrt{2meV}}
Explanation:
This formula gives the de Broglie wavelength of an electron accelerated by a potential VV.
Example:If V=100V = 100 V, h=6.63×10−34h = 6.63 \times 10^{-34} Js, and m=9.1×10−31m = 9.1 \times 10^{-31} kg:
λ=6.63×10−342×9.1×10−31×1.6×10−19×100≈1.23×10−10 m