Q: The frequency of light of wavelength 600 nm (c = 3 × 10⁸ m/s) is:

5 \times 10^{14} Hz. ν = c/λ = (3 × 10⁸)/(600 × 10⁻⁹) = 5 × 10¹⁴ Hz.

Q: The energy of a photon of frequency 5 × 10¹⁴ Hz (h = 6.626 × 10⁻³⁴ J s) is about:

3.3 \times 10^{-19} J. E = hν = (6.626 × 10⁻³⁴)(5 × 10¹⁴) ≈ 3.3 × 10⁻¹⁹ J.

Q: The energy of a photon of wavelength 400 nm (h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s) is approximately:

5.0 \times 10^{-19} J. E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(400 × 10⁻⁹) ≈ 4.97 × 10⁻¹⁹ ≈ 5.0 × 10⁻¹⁹ J.

Q: The correct order of increasing wavelength for the following radiations is:

γ-rays < X-rays < UV < Infrared. Wavelength increases (frequency decreases) in the order γ-rays < X-rays < UV < visible < Infrared.

Q: According to Planck's quantum theory, energy is emitted or absorbed only in discrete packets called:

Quanta (photons). Planck proposed that radiant energy is emitted/absorbed in whole-number multiples of a quantum, E = hν (a quantum of light is a photon).

Q: In the equation E = hν, the constant h is the:

Planck constant (6.626 × 10⁻³⁴ J s). h is Planck's constant, 6.626 × 10⁻³⁴ J s, relating a photon's energy to its frequency.

Question 1

The relationship between the speed (c), frequency (ν) and wavelength (λ) of electromagnetic radiation is:

Accepted Answer

c = νλ. For any electromagnetic wave, speed = frequency × wavelength, i.e. c = νλ.

Question 2

The wavenumber (ν̄) of radiation is defined as:

Accepted Answer

The reciprocal of wavelength (1/λ). Wavenumber ν̄ = 1/λ, the number of waves per unit length (units m⁻¹ or cm⁻¹).

Question 3

The frequency of light of wavelength 600 nm (c = 3 × 10⁸ m/s) is:

Accepted Answer

5 	imes 10^{14} Hz. ν = c/λ = (3 × 10⁸)/(600 × 10⁻⁹) = 5 × 10¹⁴ Hz.

Question 4

The energy of a photon of frequency 5 × 10¹⁴ Hz (h = 6.626 × 10⁻³⁴ J s) is about:

Accepted Answer

3.3 	imes 10^{-19} J. E = hν = (6.626 × 10⁻³⁴)(5 × 10¹⁴) ≈ 3.3 × 10⁻¹⁹ J.

Question 5

The energy of a photon of wavelength 400 nm (h = 6.626 × 10⁻³⁴ J s, c = 3 × 10⁸ m/s) is approximately:

Accepted Answer

5.0 	imes 10^{-19} J. E = hc/λ = (6.626 × 10⁻³⁴ × 3 × 10⁸)/(400 × 10⁻⁹) ≈ 4.97 × 10⁻¹⁹ ≈ 5.0 × 10⁻¹⁹ J.

Question 6

Which of the following electromagnetic radiations has the highest frequency?

Accepted Answer

γ-rays. γ-rays have the shortest wavelength and therefore the highest frequency (and energy) among the given radiations.

Question 7

Which of the following has the longest wavelength?

Accepted Answer

Radio waves. Radio waves have the longest wavelength (and lowest frequency) in the electromagnetic spectrum.

Question 8

The correct order of increasing wavelength for the following radiations is:

Accepted Answer

γ-rays < X-rays < UV < Infrared. Wavelength increases (frequency decreases) in the order γ-rays < X-rays < UV < visible < Infrared.

Question 9

According to Planck's quantum theory, energy is emitted or absorbed only in discrete packets called:

Accepted Answer

Quanta (photons). Planck proposed that radiant energy is emitted/absorbed in whole-number multiples of a quantum, E = hν (a quantum of light is a photon).

Question 10

In the equation E = hν, the constant h is the:

Accepted Answer

Planck constant (6.626 × 10⁻³⁴ J s). h is Planck's constant, 6.626 × 10⁻³⁴ J s, relating a photon's energy to its frequency.

Question 11

A perfect black body is one which:

Accepted Answer

Absorbs and emits radiation of all frequencies. An ideal black body is a perfect absorber and, when heated, a perfect emitter of radiation of all wavelengths.

Question 12

The photoelectric effect is the:

Accepted Answer

Ejection of electrons from a metal surface when light of suitable frequency strikes it. When light of frequency above a threshold hits a metal, electrons (photoelectrons) are ejected — the photoelectric effect.

Question 13

In the photoelectric effect, the kinetic energy of the emitted electrons depends on the:

Accepted Answer

Frequency of the incident light. The maximum kinetic energy of photoelectrons depends only on the frequency of the light (KEmax = hν − φ), not its intensity.

Question 14

In the photoelectric effect, the number of photoelectrons ejected per second depends on the:

Accepted Answer

Intensity of the incident light. The number of electrons ejected (hence the photocurrent) is proportional to the intensity of the light, provided ν > ν₀.

Question 15

The minimum frequency of light required to eject an electron from a metal surface is called the:

Accepted Answer

Threshold frequency. The threshold frequency ν₀ is the minimum frequency below which no photoelectrons are emitted, whatever the intensity.

Question 16

The work function (φ) of a metal is related to its threshold frequency (ν₀) by:

Accepted Answer

φ = hν₀. The work function is the minimum energy to remove an electron: φ = hν₀.

Question 17

The work function of a metal is 3.3 × 10⁻¹⁹ J. Its threshold frequency (h = 6.6 × 10⁻³⁴ J s) is:

Accepted Answer

5 	imes 10^{14} Hz. ν₀ = φ/h = (3.3 × 10⁻¹⁹)/(6.6 × 10⁻³⁴) = 5 × 10¹⁴ Hz.

Question 18

Light of frequency 1 × 10¹⁵ Hz falls on a metal of work function 3.3 × 10⁻¹⁹ J (h = 6.6 × 10⁻³⁴ J s). The maximum kinetic energy of the ejected electrons is:

Accepted Answer

3.3 	imes 10^{-19} J. KEmax = hν − φ = (6.6 × 10⁻³⁴ × 1 × 10¹⁵) − 3.3 × 10⁻¹⁹ = 6.6 × 10⁻¹⁹ − 3.3 × 10⁻¹⁹ = 3.3 × 10⁻¹⁹ J.

Question 19

If the frequency of the incident light is less than the threshold frequency of the metal, then:

Accepted Answer

No electrons are ejected no matter how intense the light. Below the threshold frequency, no photoelectron is emitted regardless of intensity — a key point the wave theory could not explain.

Question 20

The photoelectric effect was successfully explained by:

Accepted Answer

Einstein, using the particle (photon) nature of light. Einstein (1905) explained the photoelectric effect by treating light as a stream of photons of energy hν.

Developments Leading to Bohr's Model — JEE Main Chemistry MCQs with Solutions

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