Question

The energies E1 and E2 of two radiations are 2.5eV and 5.0eV the relation between there wavelengths will be

Answer 1

Answer:

E1=25eV

E2=50ev

We know that

Energy,E=xc/lambda

So,

E1/E2=lamda2/lambda1

Lambda2/lambda1=25eV/50eV

=lambda1=2*lamda2

Explanation:

Answer 2

AnswEr :

Given that,

Two radiations have energies 2.5eV and 5eV respectively.

To finD

The ratio of their wavelengths

Consider the following relation.

$\huge{\boxed{\boxed{\sf E = \dfrac{hc}{\lambda}}}}$

Here,

$\sf{Here} \begin{cases} \sf{E \longrightarrow Energy } \\ \sf{h \longrightarrow Planck's \ Constant } \\ \sf{\lambda \longrightarrow Wavelength} \\ \sf{c \longrightarrow Speed \ Of \ Light} \end{cases}$

Also,

• h = 6.626 × 10^-34 Js

• c = 3 × 10^8 m/s

Applying the Proportionality Sign,

$\sf E \propto \dfrac{1}{\lambda}$

Thus,

$\implies \sf \dfrac{E_1}{E_2} = \dfrac{\lambda_2}{\lambda_1} \\ \\ \implies \sf \dfrac{2.5}{5} = \dfrac{\lambda_2}{\lambda_1} \\ \\ \implies \huge {\sf \lambda_1 = 2\lambda_2 }$

Dual Nature of Matter :

Schrodinger hypothesised that microscopic particle open occur as both wave and particle

For example,

Photon behaves both as a wave and particle

• Wave Nature of Photon : Diffraction

• Particle Nature of Photon : Scintillation

He further derived a relation to calculate the wavelength of the wave/particle under consideration

From Theory of Relativity,

$\sf E = mc^2 ------------(1)$

Also,

$\sf E = h \nu \\ \\ \dashrightarrow \sf E = h \times \dfrac{c}{\lambda} -----------(2)$

From relations (1) and (2),

$\dashrightarrow \ \sf mc^2 = \dfrac{hc}{\lambda} \\ \\ \dashrightarrow \ \boxed{\boxed{\sf \lambda = \dfrac{h}{p}}}$