Question

calculate the frequency and wavelength of photon with the energy 3.98×10^_15J.

plzzzzz. explain in detail....plzzzzzs

Answer 1

we know
E=h×n
where E=energy
h is the plank constant=6.623×10^-34(si unit)
and n is the frequency.

so n=E÷h
=6.01×10^18 hz

again we know C=n×L
where C is the velocity of light (photon)=3×10^8 m/sec
L is the wavelength.

so, L=C÷n
=4.99×10^-11 m

Answer 2

$\text{\underline{We\:know\:that,}}$

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


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

$\boxed{E\:=\:hv}$ (Planck's quantum theory)


$\sf{\underline{Here:}}$

E = $3.98 \times 10 ^{ - 15} J$

h = $6.626 \times 10 ^{ - 34} Js$

v = ?


$\text{\underline{So,}}$

${E\:=\:hv}$

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

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

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


$\text{\underline{NOTE:}}$

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


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

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


$\sf{\underline{Here:}}$

$\lambda$ = ?

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

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


$\text{\underline{So,}}$

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

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

$\lambda = 0.5 \times {10}^{ - 10} m$

$\lambda$ = 0.5 Å


$\text{\underline{NOTE:}}$

1 Å = $\boxed{{10}^{ - 10} m}$