Question

Two of the stable isotopes of mercury are 198Hg and 202Hg. These isotopes emit slightly different wavelengths of light, 546.07532nm and 546.07355nm, respectively.

To identify which isotopes are included in a sample of mercury, we can shine the light it emits on a diffraction grating with a screen 6 m away. How many lines per centimeter are necessary in the diffraction grating for the first-order bright fringes of the two wavelengths to be 1.4 cm apart on the screen?

Answer 1

using equation

n = d sin

sin = n / d

then

sin₁ = n ₁ / d

₁ n ₁ / d

sin₂ = n ₂ / d

₂ n ₂ / d

= ₁ - ₂

= (n ₁ / d) - (n ₂ / d)

= n ( ₁ - ₂ ) / d

for first order n = 1

= ( ₁ - ₂ ) / d

d = 0.00177 x ( 6 / 0.014 )

d = 0.75857 nm

= 0.75857 x 10^-9 m

= 0.75857 x 10^-7 cm

N = 1 / d

N = 1 / 0.75857 x 10^-7

= 13182700.0257 / cm

N = 1318.27 x 10⁴ /cm

so the lines per centimeter are necessary in the diffraction grating for the

first-order bright fringes of the two wavelengths to be 1.4 cm apart on the screen is N = 1318.27 x 10⁴ /cm