Q.9
A cube of unit volume (1 m³) contains 35×107 photons of frequency 1015Hz. If the energy of all the photons is viewed as the average energy being contained in the electromagnetic waves within the same volume, then the amplitude of the magnetic field is α×109T. Given:

  • Permeability of free space μ0=4π×107Tm/A,

  • Planck’s constant h=6×1034Js,

  • π=227.

The value of α is ______.

Answer Range: [21 to 25]

1. Understanding the Problem

We are given:

  • A cube of volume V=1m3 containing N=35×107 photons.

  • Each photon has frequency ν=1015Hz.

  • The total energy of these photons is equivalent to the energy density of an electromagnetic (EM) wave in the same volume.

  • We need to find the amplitude B0 of the magnetic field in the EM wave, expressed as α×109T.

2. Calculate Total Energy of Photons

The energy of one photon is:

Ephoton=hν=6×1034×1015=6×1019J.

Total energy of all photons:

Etotal=N×Ephoton=35×107×6×1019=210×1012J.

3. Relate Photon Energy to EM Wave Energy Density

The energy density u of an EM wave is the sum of the electric and magnetic field energy densities:

u=12ϵ0E02+12μ0B02.

For an EM wave, E0=cB0, where c is the speed of light (c3×108m/s). Substituting E0:

u=12ϵ0(cB0)2+12μ0B02=B022μ0(1+μ0ϵ0c2).

Using c2=1μ0ϵ0, this simplifies to:

u=B02μ0.

The total energy in volume V=1m3 is:

Etotal=u×V=B02μ0.

Equate this to the total photon energy:

B02μ0=210×1012.

4. Solve for B0

Given μ0=4π×107Tm/A and π=227:

B02=210×1012×4π×107=840π×1019.

Substitute π=227:

B02=840×227×1019=2640×1019.

Take the square root:

B0=2640×1019=26.4×1018=26.4×109T.

Approximate 26.45.14:

B05.14×109T.

Thus, α5.14.

5. Reconcile with Expected Answer

The expected answer range is [21 to 25], but our calculation gives α5.14. This discrepancy suggests:

  • The problem might consider the peak energy density rather than the average. For a sinusoidal EM wave, the peak energy density is twice the average:

    upeak=2uavg=2×210×1012=420×1012J/m3.

    Recalculating:

    B02μ0=420×1012    B02=1680π×1019.B0=1680×227×1019=5280×10197.27×109T.

    This gives α7.27, still outside the expected range.

  • Alternatively, the problem might involve standing waves or other configurations where the energy density is higher. However, without additional information, the most accurate calculation based on the given data yields α5.14.

6. Final Answer

Based on standard EM wave theory and the given data, the value of α is:

\boxed{5.14}

(Note: The expected range [21 to 25] may imply additional context or corrections not provided in the problem statement.)

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