Question

find the energy of each of photon which have wave length of 0.50A h=6.27×10-34 js​

Answer 1

To Find :-

• The energy of a photon.

Solution :-

• Wave length (λ) = 0.50 Å
• Plank's constant (h) = 6.27 × $\sf{10^{-34}}$ Js
• Speed of light (c) = 3 ×$\sf{ 10^8}$ m/s

Photon energy formula is given by,

$\sf \red \bigstar\:\: \bold {E = \frac{hc}{\lambda}}$

[ Putting values ]

$\sf \longrightarrow \: E = \frac{6.27 \times 10 {}^{ - 34} \times 3 \times 10 {}^{8} }{0.50}$

$\sf \longrightarrow \: E = \frac{6.27 \times 3 \times 10 {}^{ - 34 + 8} }{0.50}$

$\sf \longrightarrow \: E = \frac{18.81 \times 10 {}^{ - 26} }{0.50}$

$\longrightarrow \: \bold{E = 37.62 \times 10^{-26} \: joule} \: \: \green \bigstar$

Therefore,

The energy of a photon is $\sf \: 37.62 \times 10 {}^{ - 26}J.$

Answer 2

Explanation:

Given:-

wave length =${\lambda}=0.50Å$

Constant of Plank=h=6.24×${10}^{-34}Js$

Speed of light=c ${3}×{10}^{-8}$

To find:-

The energy of photon.=E

Solution:-

As we know that

${:}\longrightarrow$${\boxed {E={\frac {hc}{{\lambda }}}}}$

${:}\longrightarrow$${\frac {6.24×{10}^{-34}×3×{10}^{-8}}{0.50}}$

${:}\longrightarrow$${\frac {6.24×3×{10}^{-34+8}}{0.50}}$

${:}\longrightarrow$${\frac {18.81×{10}^{-26}}{0.50}}$

${:}\longrightarrow$$37.62×{10}^{-26}Joule$

${:}\longrightarrow$${\underline{\boxed{\bf {E=37.62×{10}^{-26}}}}}$