Question

calculate the energy of photon corresponding to ultraviolet light and red light, give that their wavelengths are 3000A° and 7000A° respectively ​

Answer 1

Given :

• Wavelength of ultraviolet light = 3000 A⁰
• Wavelength of red light = 7000 A⁰

To find :

• The energy of photon corresponding to ultraviolet light and red light.

Solution :

The relation between wavelength and Energy is given by ,

$\boxed {\rm{E = \frac{hc}{ \lambda} }}$

Where ,

• λ is wavelength
• h is planck's constant ( = 6.625 × 10⁻³⁴ J.S)
• c is velocity of light ( = 3 × 10⁸ m/s)

The relation between Angstroms and metres is given by ,

$\boxed {\rm{1A {}^{0} = {10}^{ - 10} \: m}}$

$: \implies \rm \lambda = 3000 \times {10}^{ - 10} \: m \\ \\ : \implies \rm \lambda = 3 \times {10}^{ - 7} \: m$

We have ,

• λ = 3 × 10⁻⁷ m
• c = 3 × 10⁸ m/s
• h = 6.625 × 10⁻³⁴ J.s

Energy asscociated with the ultraviolet light is ,

$: \implies \rm \: E = \frac{(6.625 \times {10}^{ - 34}J.s)(3 \times {10}^{8} \: m {s}^{ - 1}) }{(3 \times {10}^{ - 7} \: m ) } \\ \\ : \implies \rm \: E = \frac{(6.625 \times 10 {}^{ - 34} \: J \: \cancel{s}) \times (3 \times {10}^{8} \cancel{m {s}^{ - 1} }) }{(3 \times {10}^{ - 7} \cancel{m}) } \\ \\ : \implies \rm \: E = \frac{6.625 \times {10}^{ - 34} J \times 3 \times 10 {}^{8} \times {10}^{7} }{3} \\ \\ : \implies \rm \: E= \frac{19.875 \times 10 {}^{ - 34} \times {10}^{15} \: J}{3} \\ \\ : \implies \rm \: E = 6.625 \times {10}^{ - 19} \: J$

Hence , The Energy associated with the ultraviolet light is 6.625 × 10⁻¹⁹ J

Now , wavelength of red light is

$: \implies \rm \lambda = 7000 \times {10}^{ - 10} \: m \\ \\ : \implies \rm \lambda = 7 \times {10}^{ - 7} \: m$

The Energy associated with the red light is ,

$: \implies \rm \: E = \frac{(6.625 \times {10}^{ - 34}J.s)(3 \times {10}^{8} \: m {s}^{ - 1}) }{(7 \times {10}^{ - 7} \: m ) } \\ \\ : \implies \rm \: E = \frac{(6.625 \times 10 {}^{ - 34} \: J \: \cancel{s}) \times (3 \times {10}^{8} \cancel{m {s}^{ - 1} }) }{(7 \times {10}^{ - 7} \cancel{m}) } \\ \\ : \implies \rm \: E = \frac{6.625 \times {10}^{ - 34} J \times 3 \times 10 {}^{8} \times {10}^{7} }{7} \\ \\ : \implies \rm \: E = \frac{19.875 \times 10 {}^{ - 34} \times {10}^{15} \: J}{7} \\ \\ : \implies \rm \: E = 2.83\times {10}^{ - 19} \: J$

Hence , The Energy associated with the red light is 2.83 × 10⁻¹⁹ J .

Answer 2

Answer:

Given :

Wavelength of ultraviolet light = 3000 A⁰

Wavelength of red light = 7000 A⁰

To find :

The energy of photon corresponding to ultraviolet light and red light.

Solution :

The relation between wavelength and Energy is given by ,

\boxed {\rm{E = \frac{hc}{ \lambda} }}

E=

λ

hc

Where ,

λ is wavelength

h is planck's constant ( = 6.625 × 10⁻³⁴ J.S)

c is velocity of light ( = 3 × 10⁸ m/s)

The relation between Angstroms and metres is given by ,

\boxed {\rm{1A {}^{0} = {10}^{ - 10} \: m}}

1A

0

=10

−10

m

\begin{gathered} : \implies \rm \lambda = 3000 \times {10}^{ - 10} \: m \\ \\ : \implies \rm \lambda = 3 \times {10}^{ - 7} \: m\end{gathered}

:⟹λ=3000×10

−10

m

:⟹λ=3×10

−7

m

We have ,

λ = 3 × 10⁻⁷ m

c = 3 × 10⁸ m/s

h = 6.625 × 10⁻³⁴ J.s

Energy asscociated with the ultraviolet light is ,

\begin{gathered} : \implies \rm \: E = \frac{(6.625 \times {10}^{ - 34}J.s)(3 \times {10}^{8} \: m {s}^{ - 1}) }{(3 \times {10}^{ - 7} \: m ) } \\ \\ : \implies \rm \: E = \frac{(6.625 \times 10 {}^{ - 34} \: J \: \cancel{s}) \times (3 \times {10}^{8} \cancel{m {s}^{ - 1} }) }{(3 \times {10}^{ - 7} \cancel{m}) } \\ \\ : \implies \rm \: E = \frac{6.625 \times {10}^{ - 34} J \times 3 \times 10 {}^{8} \times {10}^{7} }{3} \\ \\ : \implies \rm \: E= \frac{19.875 \times 10 {}^{ - 34} \times {10}^{15} \: J}{3} \\ \\ : \implies \rm \: E = 6.625 \times {10}^{ - 19} \: J\end{gathered}

:⟹E=

(3×10

−7

m)

(6.625×10

−34

J.s)(3×10

8

ms

−1

)

:⟹E=

(3×10

−7

m

)

(6.625×10

−34

J

s

)×(3×10

8

ms

−1

)

:⟹E=

3

6.625×10

−34

J×3×10

8

×10

7

:⟹E=

3

19.875×10

−34

×10

15

J

:⟹E=6.625×10

−19

J

Hence , The Energy associated with the ultraviolet light is 6.625 × 10⁻¹⁹ J

Now , wavelength of red light is

\begin{gathered} : \implies \rm \lambda = 7000 \times {10}^{ - 10} \: m \\ \\ : \implies \rm \lambda = 7 \times {10}^{ - 7} \: m\end{gathered}

:⟹λ=7000×10

−10

m

:⟹λ=7×10

−7

m

The Energy associated with the red light is ,

\begin{gathered} : \implies \rm \: E = \frac{(6.625 \times {10}^{ - 34}J.s)(3 \times {10}^{8} \: m {s}^{ - 1}) }{(7 \times {10}^{ - 7} \: m ) } \\ \\ : \implies \rm \: E = \frac{(6.625 \times 10 {}^{ - 34} \: J \: \cancel{s}) \times (3 \times {10}^{8} \cancel{m {s}^{ - 1} }) }{(7 \times {10}^{ - 7} \cancel{m}) } \\ \\ : \implies \rm \: E = \frac{6.625 \times {10}^{ - 34} J \times 3 \times 10 {}^{8} \times {10}^{7} }{7} \\ \\ : \implies \rm \: E = \frac{19.875 \times 10 {}^{ - 34} \times {10}^{15} \: J}{7} \\ \\ : \implies \rm \: E = 2.83\times {10}^{ - 19} \: J\end{gathered}

:⟹E=

(7×10

−7

m)

(6.625×10

−34

J.s)(3×10

8

ms

−1

)

:⟹E=

(7×10

−7

m

)

(6.625×10

−34

J

s

)×(3×10

8

ms

−1

)

:⟹E=

7

6.625×10

−34

J×3×10

8

×10

7

:⟹E=

7

19.875×10

−34

×10

15

J

:⟹E=2.83×10

−19

J

Hence , The Energy associated with the red light is 2.83 × 10⁻¹⁹ J .