Question

calculate
the frequency and wavelength of photon with energy 3.98 into 10 ki power minus 15 ​

Answer 1

Answer:

Weknowthat,

Energy of photon = \sf\boxed{3.98 \times {10}^{ - 15} \: J}

3.98×10

−15

J

(a) \text{\underline{Frequency\:of\:photon:}}

Frequencyofphoton:

\boxed{E\:=\:hv}

E=hv

(Planck's quantum theory)

\sf{\underline{Here:}}

Here:

E = 3.98 \times 10 ^{ - 15} J3.98×10

−15

J

h = 6.626 \times 10 ^{ - 34} Js6.626×10

−34

Js

v = ?

\text{\underline{So,}}

So,

{E\:=\:hv}E=hv

3.98 \times 10 ^{ - 15} J = 6.626 \times 10 ^{ - 34} Js \times v3.98×10

−15

J=6.626×10

−34

Js×v

v = \frac{3.98 \times 10 ^{ - 15 } J}{6.626 \times {10}^{ - 34} Js}v=

6.626×10

−34

Js

3.98×10

−15

J

\sf\boxed{Frequency = 6.0 \times {10}^{18} s ^{ - 1} (Hz)}

Frequency=6.0×10

18

s

−1

(Hz)

\text{\underline{NOTE:}}

NOTE:

Energy (E) of a photon is given by E = hv, where h is Planck's constant and v is the frequency of photon.

(b) \text{\underline{Wavelength\:of\:photon:}}

Wavelengthofphoton:

\boxed{\lambda = \frac{c}{v}}

λ=

v

c

\sf{\underline{Here:}}

Here:

\lambdaλ = ?

c = {3 \times {10}^{8}ms ^{ - 1} }3×10

8

ms

−1

v = {6.0 \times {10}^{18} {s}^{ - 1} }6.0×10

18

s

−1

\text{\underline{So,}}

So,

\lambda = {\frac{c}{v}}λ=

v

c

\lambda = \frac{3 \times {10}^{8}ms ^{ - 1} }{6.0 \times {10}^{18} {s}^{ - 1} }λ=

6.0×10

18

s

−1

3×10

8

ms

−1

\lambda = 0.5 \times {10}^{ - 10} mλ=0.5×10

−10

m

\lambdaλ = 0.5 Å

\text{\underline{NOTE:}}

NOTE:

1 Å = \boxed{{10}^{ - 10} m}

10

−10

m