The work functions of two metals (MA and MB) are in the 1:2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of MA:MB is in the ratio of 2.642:1. The work functions (in eV) of MA and MB are respectively.

where E is photon energy and phi is the work function. Given the ratio of work functions is 1:2, let the work function of MA be phi1 and MB be phi2. Then phi2 = 2phi1. For metal MA exposed to 6 eV photons: (KEmax)1 = 6 - phi1 --- (Equation 1) For metal MB exposed to 6 eV photons: (KEmax)2 = 6 - phi2 = 6 - 2phi1 --- (Equation 2) We are given the kinetic energy ratio: Substitute equations 1 and 2 into the ratio: Since phi2 = 2phi1: Photoelectric effect ratio problems are purely algebraic setups. Express all unknowns in terms of a single variable using the given multiplier (1:2 ratio) and plug directly into Einstein's photoelectric equation. Class 11 Chemistry: Structure of Atom Class 12 Physics: Dual Nature of Radiation and Matter

The work functions of two metals (MA and MB) are in the 1:2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of MA:MB is in the ratio of 2.642:

where E is photon energy and phi is the work function. Given the ratio of work functions is 1:2, let the work function of MA be phi1 and MB be phi2. Then phi2 = 2phi1. For metal MA exposed to 6 eV photons: (KEmax)1 = 6 - phi1 --- (Equation 1) For metal MB exposed to 6 eV photons: (KEmax)2 = 6 - phi2 = 6 - 2phi1 --- (Equation 2) We are given the kinetic energy ratio: Substitute equations 1 and 2 into the ratio: Since phi2 = 2phi1: Photoelectric effect ratio problems are purely algebraic setups. E

The work functions of two metals (MA and MB) are in the 1:2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of MA:MB is in the ratio of 2.642:1. The work functions (in eV) of MA and MB are respectively.

where E is photon energy and phi is the work function. Given the ratio of work functions is 1:2, let the work function of MA be phi1 and MB be phi2. Then phi2 = 2phi1. For metal MA exposed to 6 eV photons: (KEmax)1 = 6 - phi1 --- (Equation 1) For metal MB exposed to 6 eV photons: (KEmax)2 = 6 - phi2 = 6 - 2phi1 --- (Equation 2) We are given the kinetic energy ratio: Substitute equations 1 and 2 into the ratio: Since phi2 = 2phi1: Photoelectric effect ratio problems are purely algebraic setups. Express all unknowns in terms of a single variable using the given multiplier (1:2 ratio) and plug directly into Einstein's photoelectric equation. Class 11 Chemistry: Structure of Atom Class 12 Physics: Dual Nature of Radiation and Matter

The work functions of two metals (MA and MB) are in the 1:2 ratio. When these metals are exposed to photons of energy 6 eV, the kinetic energy of liberated electrons of MA:MB is in the ratio of 2.642:

where E is photon energy and phi is the work function. Given the ratio of work functions is 1:2, let the work function of MA be phi1 and MB be phi2. Then phi2 = 2phi1. For metal MA exposed to 6 eV photons: (KEmax)1 = 6 - phi1 --- (Equation 1) For metal MB exposed to 6 eV photons: (KEmax)2 = 6 - phi2 = 6 - 2phi1 --- (Equation 2) We are given the kinetic energy ratio: Substitute equations 1 and 2 into the ratio: Since phi2 = 2phi1: Photoelectric effect ratio problems are purely algebraic setups. E

The work functions of two metals (MA and MB) are in the 1:2 ratio. When these metals are exposed to photons of energy 6…

Year	2026	2025	2024	Total
Questions	14	16	9	39

Q70 jee_main_2026_21_jan_morning Hydrogen Spectrum

Given below are two statements: Statement I: When an electric discharge is passed through gaseous hydrogen, the hydrogen molecules dissociate and the energetically excited hydrogen atoms produce electromagnetic radiation of discrete frequencies. Statement II: The frequency of second line of Balmer series obtained from

He⁺

is equal to that of first line of Lyman series obtained from hydrogen atom. In the light of the above statements, choose the correct answer from the options given below:

A. $Both Statement I and Statement II are true$
B. $Both Statement I and Statement II are false$
C. $Statement I is false but Statement II is true$
D. $Statement I is true but Statement II is false$

Solution

Related Formula

(1)/(λ) = RZ²((1)/(n₁²) - (1)/(n₂²))

Core Logic

Statement I is a factual description of how the hydrogen emission spectrum is obtained. Dissociation yields excited atoms that emit light at discrete frequencies. So Statement I is true.

For Statement II: First line of Lyman series for H atom ( $Z=1, n₁=1, n₂=2$ $Z=1, n_1=1, n_2=2$):

(1)/(λ₁) = R(1)²((1)/(1²) - (1)/(2²)) = R(1 - (1)/(4)) = (3R)/(4)

Second line of Balmer series for $He^+$ $He^+$ ( $Z=2, n₁=2, n₂=4$ $Z=2, n_1=2, n_2=4$):

(1)/(λ₂) = R(2)²((1)/(2²) - (1)/(4²)) = 4R((1)/(4) - (1)/(16)) = 4R((3)/(16)) = (3R)/(4)

Since $(1)/(λ₁) = (1)/(λ₂)$ , their wavelengths (and therefore frequencies) are exactly the same. Statement II is true.

Step 1: Final Conclusion

Both statements are correct.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q51 jee_main_2026_21_jan_evening Hydrogen Spectrum and Energy Levels

Consider the following spectral lines for atomic hydrogen: A. First line of Paschen series B. Second line of Balmer series C. Third line of Paschen series D. Fourth line of Bracket series The correct arrangement of the above lines in ascending order of energy is:

A. $(1) D < C < A < B$ $(1) \ D < C < A < B$
B. $(2) A < B < C < D$ $(2) \ A < B < C < D$
C. $(3) C < D < B < A$ $(3) \ C < D < B < A$
D. $(4) D < A < C < B$ $(4) \ D < A < C < B$

Solution

Related Formula

Δ E = 13.6 Z² ((1)/(n₁²) - (1)/(n₂²)) eV

Core Logic

Let's find the values of $n₁$ $n_1$ and $n₂$ $n_2$ for each transition:

(A) Paschen ( $1st$ $1^{st}$ line): $n₁ = 3, n₂ = 4$ $n_1 = 3, n_2 = 4$
(B) Balmer ( $2nd$ $2^{nd}$ line): $n₁ = 2, n₂ = 4$ $n_1 = 2, n_2 = 4$
(C) Paschen ( $3rd$ $3^{rd}$ line): $n₁ = 3, n₂ = 6$ $n_1 = 3, n_2 = 6$
(D) Bracket ( $4th$ $4^{th}$ line): $n₁ = 4, n₂ = 8$ $n_1 = 4, n_2 = 8$

Calculating or comparing the energy values corresponding to these transitions yields the ascending order of energy.

Step 1: Final Conclusion

The correct ascending order of energy of the given lines is $D < A < C < B$ $D < A < C < B$, corresponding to option (4).

Pattern Recognition

Sees: hydrogen spectral lines energy comparison. Trap: Confusing series limits with specific line numbers. Shortcut: Evaluate transition frequencies or wavelength gaps using Rydberg formula equivalents.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q54 jee_main_2026_22_january_morning Bohr Model Energy Calculations

The energy required by electrons, present in the first Bohr orbit of hydrogen atom to be excited to second Bohr orbit is ____

J mol⁻¹

. Given:

RH = 2.18 \times 10⁻¹¹ ergs

.

A. $1.635 \times 10⁻¹⁸$ $1.635 \times 10^{-18}$
B. $9.835 \times 10⁵$ $9.835 \times 10^{5}$
C. $9.835 \times 10¹²$ $9.835 \times 10^{12}$
D. $1.635 \times 10⁻¹¹$ $1.635 \times 10^{-11}$

Solution

Related Formula

Eₙ = -RH \times Z²n²

Δ E = RH Z² ( (1)/(n₁²) - (1)/(n₂²) )

Core Logic

Given $RH = 2.18 \times 10⁻¹¹ ergs$ . Convert this to Joules: $1 Joule = 10⁷ ergs RH = 2.18 \times 10⁻¹⁸ J$ .

Calculate energy difference per atom:

Δ E = 2.18 \times 10⁻¹⁸ \times 1² [ 11² - 12² ]

Δ E = 2.18 \times 10⁻¹⁸ \times ( 1 - (1)/(4) ) = 2.18 \times 10⁻¹⁸ \times (3)/(4)

Δ E = 1.635 \times 10⁻¹⁸ Joule/atom

Step 1: Conversion to per mole

To find the energy per mole, multiply by Avogadro's number ( $NA = 6.02 \times 10²³$ $N_A = 6.02 \times 10^{23}$):

Δ Emole = 1.635 \times 10⁻¹⁸ \times 6.02 \times 10²³ Joule/mole

Δ Emole = 9.84 \times 10⁵ Joule/mole \approx 9.835 \times 10⁵ J mol⁻¹

Pattern Recognition

Energy gaps in Hydrogen: $1 arrow 2$ $1 \rightarrow 2$ transition is exactly $(3)/(4)$ $\frac{3}{4}$ of the ionization energy. Watch out for per atom vs per mole unit traps.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Q64 jee_main_2026_22_january_evening Balmer Series Energy Transitions

The energy of first (lowest) Balmer line of H atom is x J. The energy (in J) of second Balmer line of H atom is:

A. $x²$ $x^2$
B. $(x)/(1.35)$ $\frac{x}{1.35}$
C. $2x$ $2x$
D. $1.35x$ $1.35x$

Solution

Related Formula

Δ E = 13.6 Z² ((1)/(n₁²) - (1)/(n₂²)) eV

Core Logic

Step 1: First Balmer line ( $n₁ = 2, n₂ = 3$ $n_1 = 2, n_2 = 3$):

Δ E₁ = x = 13.6 ((1)/(2²) - (1)/(3²)) = 13.6 ((1)/(4) - (1)/(9)) = 13.6 ((5)/(36))

Step 2: Second Balmer line ( $n₁ = 2, n₂ = 4$ $n_1 = 2, n_2 = 4$):

Δ E₂ = 13.6 ((1)/(2²) - (1)/(4²)) = 13.6 ((1)/(4) - (1)/(16)) = 13.6 ((3)/(16))

Step 3: Ratio of energies:

(Δ E₂)/(x) = ((3)/(16))/((5)/(36)) = (3 \times 36)/(16 \times 5) = (27)/(20) = 1.35

Δ E₂ = 1.35 x

Pattern Recognition

Sees: Ratio of hydrogen spectrum spectral line energies. Shortcut: $(3/16) / (5/36) = 27/20 = 1.35$ $(3/16) / (5/36) = 27/20 = 1.35$.

Chapter Mix

Class 11 Chemistry: Structure of Atom Class 12 Physics: Atoms

Q51 jee_main_2026_23_january_morning Bohr Model for Hydrogen Like Species

Which of the following statements regarding the energy of the stationary state is true in the following one-electron system?

A. $-1.09 \times 10⁻¹⁸ J for second orbit of H atom.$
B. $+2.18 \times 10⁻¹⁸ J for second orbit of He⁺ ion$
C. $+8.72 \times 10⁻¹⁸ J for first orbit of He⁺ ion$
D. $-2.18 \times 10⁻¹⁸ J for third orbit of Li²⁺ ion$

Solution

Related Formula

Eₙ = -2.18 \times 10⁻¹⁸ (Z²)/(n²) J/atom

Core Logic

Evaluate the energy of the stationary state for the given species by substituting the atomic number $Z$ $Z$ and the orbit number $n$ $n$ into the energy formula for hydrogen-like species.

Step 1: Calculation for Li2+ Ion

For the $3rd$ $3^{\text{rd}}$ orbit of the $Li²⁺$ $Li^{2+}$ ion, we have $Z = 3$ $Z = 3$ (since lithium has 3 protons) and $n = 3$ $n = 3$.

E₃ = -2.18 \times 10⁻¹⁸ \times (3²)/(3²)

E₃ = -2.18 \times 10⁻¹⁸ J

Pattern Recognition

When $Z = n$ $Z = n$, the $Z²/n²$ $Z^2/n^2$ ratio becomes $1$ $1$, immediately resulting in the ground state energy of a hydrogen atom ( $-2.18 \times 10⁻¹⁸$ $-2.18 \times 10^{-18}$ J). This is a common shortcut for identifying correct energy states.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Solution & Explanation

Related Formula

Core Logic

Step 1: Setting up the ratio

Step 2: Solving for work function

Pattern Recognition

Chapter Mix

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