Question

calculate the frequency and energy of a photon of radiation having wave length 6000A°

Answer 1

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Answer 2

$\huge{\bold {\underline {\underline{Given : - }}}}$

• Wavelength of photon = 6000A°

$\huge {\bold {\underline {\underline{To \: find : - }}}}$

• Frequency and energy of the photon

$\huge {\bold {\underline {\underline{Solution : - }}}}$

$\large {\underline{\bold {\boxed{ \bigstar{ \: | 1A {}^{0} = 10 {}^{ - 10} m}}}}}$

We have ,

• λ = 6000A° = 6000 × 10⁻¹⁰ m = 6 × 10⁻⁷ m

Relation between wavelength and frequency is given by ,

$\large {\underline {\bold {\boxed{ \bigstar{ \: | \nu = \dfrac{c}{ \lambda} }}}}}$

Where ,

• c is velocity of light
• λ is wavelength
• υ is frequency

$: \implies \nu = \dfrac{ 3 \times 10 {}^{8} }{6 \times 10 {}^{ - 7} } \\ \\ : \implies \nu \: = \dfrac{1}{2} \times 10 {}^{15} \\ \\ : \implies \nu= 0.5 \times 10 {}^{15} \\ \\ :\implies \nu = 5 \times {10}^{14}$

Relation between Energy and frequency is given by ,

$\large {\underline {\bold {\boxed{ \bigstar{ \: | E = h \nu}}}}}$

Where ,

• E is energy
• h is planck's constant
• υ is frequency

$: \implies \: E = 6.625\times 10 {}^{ - 34} \times 5 \times 10 {}^{14} \\ \\ : \implies \: E = 33.125 \times 10 {}^{ - 20} \: J$

∴ The Energy and wavelength of the photon are 33.125 × 10⁻²⁰J and 5 × 10¹⁴ s⁻¹

$\huge {\bold {\underline {\underline{Additional\:Info : - }}}}$

➣ $\large\bold{\purple{\underline{\underline{Frequency:-}}}}$

The Number of waves that pass through a given point in one sec is called frequency .It is denoted by υ . units of frequency are cycles per second , S⁻¹ , HZ

➣ $\large\bold{\green{\underline{\underline{Wavelength:-}}}}$

The distance between two successive crusts (or) troughs between two consecutive identical points in the same phase of a wave is called wavelength.units of wavelength are A° , nm , mμ , μ , Pm ,...