If the energy of a photon corresponding to a wavelength of 6000 A is 3-32 - 10-19 J- the photon energy for a wavelength of 4000 A will beA- 1-4 eVB- 3-1 eVC- 4-9 eVD- 1-6 eV

The correct option is B 3.1 eV E = hc/λ⇒E_1/E_2=λ_1/λ_2⇒3.32 × 10^ 19/E_2= 4000/6000 ⇒ E_2 =4.98 × 10^ 19J = 3.1eV ...

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Standard XII

Physics

Stopping Potential Revisited

If the energy...

Question

If the energy of a photon corresponding to a wavelength of $6000 \circ A i s 3.32 \times 10^{- 19} J$ , the photon energy for a wavelength of $4000 \circ A$ will be

1.4 eV

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4.9 eV

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3.1 eV

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1.6 eV

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Solution

The correct option is C
3.1 eV

$E = \frac{h c}{λ} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{λ_{1}}{λ_{2}} \Rightarrow \frac{3.32 \times 10^{- 19}}{E_{2}} = \frac{4000}{6000}$
$\Rightarrow E_{2} = 4.98 \times 10^{- 19} J = 3.1 e V$

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If the energy of a photon corresponding to a wavelength of 6000∘A is 3.32×10−19J, the photon energy for a wavelength of 4000∘A will be

The correct option is C 3.1 eV E=hcλ⇒E1E2=λ1λ2⇒3.32×10−19E2=40006000 ⇒E2=4.98×10−19J=3.1eV

If the energy of a photon corresponding to a wavelength of $6000 \circ A i s 3.32 \times 10^{- 19} J$ , the photon energy for a wavelength of $4000 \circ A$ will be

The correct option is C
3.1 eV

$E = \frac{h c}{λ} \Rightarrow \frac{E_{1}}{E_{2}} = \frac{λ_{1}}{λ_{2}} \Rightarrow \frac{3.32 \times 10^{- 19}}{E_{2}} = \frac{4000}{6000}$
$\Rightarrow E_{2} = 4.98 \times 10^{- 19} J = 3.1 e V$