If h is Planck's constant, the momentum of a photon of wavelength 0.01 mathringA is

Question

If h is Planck's constant, the momentum of a photon of wavelength 0.01 mathringA is (A) 10-2 h (B) h (C) 102 h (D) 1012 h

Tardigrade Admin · Accepted Answer

Momentum of photon, p=(E/c)=(h v/c) Where E is the energy of a photon and c is the velocity of light. ∴ p =(h c/c λ) [∵ v=(c/λ)] =(h/λ)=(h/0.01 × 10-10)=1012 h

Q. If $h$ is Planck's constant, the momentum of a photon of wavelength $0.01 \overset{˚}{A}$ is

$1 0^{- 2} h$

$h$

$1 0^{2} h$

$1 0^{12} h$

Momentum of photon, $p = \frac{E}{c} = \frac{h v}{c}$
Where $E$ is the energy of a photon and $c$ is the velocity of light.
$∴ p = \frac{h c}{c λ} [∵ v = \frac{c}{λ}]$
$= \frac{h}{λ} = \frac{h}{0.01 \times 1 0 ^{- 10}} = 1 0^{12} h$

Q. If h is Planck's constant, the momentum of a photon of wavelength 0.01A˚ is

10−2h

h

102h

1012h

Momentum of photon, p=cE​=chv​ Where E is the energy of a photon and c is the velocity of light. ∴p=cλhc​[∵v=λc​] =λh​=0.01×10−10h​=1012h

Q. If $h$ is Planck's constant, the momentum of a photon of wavelength $0.01 \overset{˚}{A}$ is

$1 0^{- 2} h$

$h$

$1 0^{2} h$

$1 0^{12} h$

Momentum of photon, $p = \frac{E}{c} = \frac{h v}{c}$
Where $E$ is the energy of a photon and $c$ is the velocity of light.
$∴ p = \frac{h c}{c λ} [∵ v = \frac{c}{λ}]$
$= \frac{h}{λ} = \frac{h}{0.01 \times 1 0 ^{- 10}} = 1 0^{12} h$