Question

determine energy and wave length of radiation having frequency 1.5×10 -5 Hz​

Answer 1

Given :

• Frequency of a radiation = 1.5 × 10⁻⁵ Hz

To Find :

• The energy and wavelength of the radiation

Solution :

The relation between Wavelength and frequency of a radiation is given by ,

$\\ \star \: {\boxed{\sf{\purple{ \upsilon = \dfrac{c}{ \lambda} }}}}$

Where ,

• c is velocity of light
• λ is wavelength

We have ,

• c = 3 × 10⁸ m/s
• υ = 1.5 × 10⁻⁵ Hz = 1.5 × 10⁻⁵ s⁻¹

Substituting the values in the formulae ,

$\\ : \implies \sf 1.5 \times {10}^{ - 5} {s}^{ - 1} = \dfrac{3 \times {10}^{8} \: m {s}^{ - 1} }{ \lambda}$

$\\ : \implies \sf \lambda \times 1.5 \times {10}^{ - 5} \: {s}^{ - 1} = 3 \times {10}^{8} \: m {s}^{ - 1}$

$\\ : \implies \sf \lambda = \dfrac{3 \times {10}^{8} \: m {s}^{ - 1} }{ 1.5 \times {10}^{ - 5} \: s {}^{ - 1} }$

$\\ : \implies \sf \lambda = 2 \times {10}^{8 + 5} \: m$

$\\ : \implies{\underline{\boxed{\pink{\mathfrak {\lambda =2 \times {10}^{13} \: m }}}}} \: \bigstar$

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━

The Energy of a radiation is given by ,

$\\ \star \: {\boxed{\purple{\sf{E = h \upsilon}}}}$

Where ,

• h is planck's constant (6.625 × 10⁻³⁴ J.s)
• υ is Frequency

We have ,

• h = 6.625 × 10⁻³⁴ J.s
• υ = 1.5 × 10⁻⁵ Hz = 1.5 × 10⁻⁵ s⁻¹

Substituting the values ,

$\\ : \implies \sf \: E = (6.625 \times {10}^{ - 34} \: j.s)(1.5 \times {10}^{ - 5} \: {s}^{ - 1} )$

$\\ : \implies{\underline{\boxed{\pink{\mathfrak{E = 9.93 \times {10}^{ - 39} \: J}}}}}\:\bigstar$

Hence ,

• The Energy and wavelength of the given radiation are 9.93 × 10⁻³⁹ J and 2 × 10¹³ m

Answer 2

Answer:

Given :

Frequency of a radiation = 1.5 × 10⁻⁵ Hz

To Find :

The energy and wavelength of the radiation

Solution :

The relation between Wavelength and frequency of a radiation is given by ,

\begin{gathered} \\ \star \: {\boxed{\sf{\purple{ \upsilon = \dfrac{c}{ \lambda} }}}}\end{gathered}

⋆

υ=

λ

c

Where ,

c is velocity of light

λ is wavelength

We have ,

c = 3 × 10⁸ m/s

υ = 1.5 × 10⁻⁵ Hz = 1.5 × 10⁻⁵ s⁻¹

Substituting the values in the formulae ,

\begin{gathered} \\ : \implies \sf 1.5 \times {10}^{ - 5} {s}^{ - 1} = \dfrac{3 \times {10}^{8} \: m {s}^{ - 1} }{ \lambda} \\ \\ \end{gathered}

:⟹1.5×10

−5

s

−1

=

λ

3×10

8

ms

−1

\begin{gathered} \\ : \implies \sf \lambda \times 1.5 \times {10}^{ - 5} \: {s}^{ - 1} = 3 \times {10}^{8} \: m {s}^{ - 1} \\ \\ \end{gathered}

:⟹λ×1.5×10

−5

s

−1

=3×10

8

ms

−1

\begin{gathered} \\ : \implies \sf \lambda = \dfrac{3 \times {10}^{8} \: m {s}^{ - 1} }{ 1.5 \times {10}^{ - 5} \: s {}^{ - 1} } \\ \\ \end{gathered}

:⟹λ=

1.5×10

−5

s

−1

3×10

8

ms

−1

\begin{gathered} \\ : \implies \sf \lambda = 2 \times {10}^{8 + 5} \: m \\ \\ \end{gathered}

:⟹λ=2×10

8+5

m

\begin{gathered} \\ : \implies{\underline{\boxed{\pink{\mathfrak {\lambda =2 \times {10}^{13} \: m }}}}} \: \bigstar \\ \\ \end{gathered}

:⟹

λ=2×10

13

m

★

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━━━━━

The Energy of a radiation is given by ,

\begin{gathered} \\ \star \: {\boxed{\purple{\sf{E = h \upsilon}}}}\end{gathered}

⋆

E=hυ

Where ,

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

υ is Frequency

We have ,

h = 6.625 × 10⁻³⁴ J.s

υ = 1.5 × 10⁻⁵ Hz = 1.5 × 10⁻⁵ s⁻¹

Substituting the values ,

\begin{gathered} \\ : \implies \sf \: E = (6.625 \times {10}^{ - 34} \: j.s)(1.5 \times {10}^{ - 5} \: {s}^{ - 1} ) \\ \\ \end{gathered}

:⟹E=(6.625×10

−34

j.s)(1.5×10

−5

s

−1

)

\begin{gathered} \\ : \implies{\underline{\boxed{\pink{\mathfrak{E = 9.93 \times {10}^{ - 39} \: J}}}}}\:\bigstar\\ \\\end{gathered}

:⟹

E=9.93×10

−39

J

★

Hence ,

The Energy and wavelength of the given radiation are 9.93 × 10⁻³⁹ J and 2 × 10¹³ m