The de Broglie wavelength of an oxygen molecule at 27°C$27^{\circ}C$ is x × 10⁻¹²$x \times 10^{-12}$ m. The value of x is (take Planck's constant = 6.63 × 10⁻³⁴ J.s$6.63 \times 10^{-34}\text{ J.s}$, Boltzmann constant = 1.38 × 10⁻²³ J/K$1.38 \times 10^{-23}\text{ J/K}$, mass of oxygen. Molecule = 5.31 × 10⁻²⁶ kg$5.31 \times 10^{-26}\text{ kg}$).

Related Formula

Core Logic

Using Einstein's Photoelectric equation for the two cases:

Case 1 (Wavelength λ$\lambda$, Stopping Potential 8 V$8\mathrm{\,V}$):

Case 2 (Wavelength 3λ$3\lambda$, Stopping Potential 2 V$2\mathrm{\,V}$):

Step 2: Solving the Equations

Multiply equation (ii) by 4 to eliminate e$e$:

Equating this to equation (i):

Divide entirely by hc$hc$:

Chapter Mix

Class 12 Physics: Dual Nature Of Radiation And Matter