Question

need a correct.

Hoping for a grt ans :)​

Answer 1

Correct Option Is A

Hers Is ayour Answer.

hope This Help you.

Answer 2

${\boxed{\underline{\tt{\orange{Required \: Answer:-}}}}}$

◉ Formula used:-

• E = hv ( h is plank's constant, v is frequency of proton)

• v = C/λ (C is the velocity of light, λ is the wavelength)

• Einstein equation:- Energy of photon = Kinetic energy - work function

$\sf{ K_e = E - \phi_0}$

• $\sf{K_e = \frac{1}{2} m {v}^{2}}$

◉ SOLUTION:-

☆ E = hv

or we can write it as :-

$\sf{E = \frac{hC}{ λ} (v= \frac{C}{λ} )}$

☆ Kinetic energy is given by:-

$\sf{K_e = \frac{1}{2} m {v}^{2}}$

Or

$\sf{ K_e = E - \phi_0} \: \: \: \: \: \: \: \: \: \: \: (g)$

we can write kinetic energy in terms of momentum by multiplying both numerator and denominator by m.

$⇝ \sf{K_e = \frac{1}{2} \frac{m {v}^{2} \times m}{m \times m} }$

momentum is denoted by p, and

p = mv

substitute the value of p in the expression of kinetic energy.

$⇝ \sf{K_e = \frac{1}{2} \frac{ {p}^{2} }{ m} } \: \: \: \: \: \: \: \: \: \: \: \: \: \: (1)$

momentum multiplied by velocity gives us energy:-

E = pC

$\therefore pC = \frac{hC}{λ}$

now, substitute the value of p in equation 1

$⇝ \sf{K_e = \frac{1}{2} \frac{ {h}^{2} }{mλ {d}^{2} } } \: \: \: \: \: \: \: \: \: \: \: \: \: (2)$

Equating equation (g.) Ans (2.)

$\sf{⇝ \frac{1}{2} \frac{ {h}^{2} }{m λ {d}^{2} } = E - \phi_0} \: \: \: \: \: \: \: \: \: \: \: \: (3) \\ \\ ⇝ \sf{ \frac{1}{2} \frac{ {h}^{2} }{m λ {d}^{2}} = \frac{hC}{λ} }$

multiplying and dividing LHS and RHS by wavelengths and differentiating both LHS and RHS by wavelengths

$⇝ \sf{( \frac{1}{2} \frac{ {h}^{2} }{m λ {d}^{3}} )dλ_d = ( \frac{hC}{ {λ}^{2} })dλ } \\ \\ ⇝ \sf{ \frac{d λ_d}{d \lambda} = \pink{\frac{ \lambda {d}^{3} }{ { \lambda}^{2} } }}$

So, the first option is correct