Question 364: Dual Nature of Matter and Radiation

Question

Light of wavelength $\lambda$ strikes a photoelectric surface and electrons are ejected with kinetic energy $K$. If $K$ is to be increased to exactly twice its original value, the wavelength must be changed to $\lambda^{\prime}$, such that

Accepted Answer

$\frac{h c}{\lambda}=E+\phi_0$
$\frac{h c}{\lambda^{\prime}}=2 E+\phi_0$
Dividing $\frac{\lambda^{\prime}}{\lambda}=\left(\frac{E+\phi_0}{2 E+\phi_0}ight)$ or $\frac{\lambda^{\prime}}{\lambda}\lambda^{\prime}\quad 	ext{.... (i)}$
Also, $\quad \frac{\lambda^{\prime}}{\lambda}=\frac{1}{2}\left[\frac{E+\phi_0}{E+\frac{\phi_0}{2}}ight]$
or $\quad \frac{\lambda^{\prime}}{\lambda}>\frac{1}{2}$ or $\lambda^{\prime}>\frac{\lambda}{2}\quad 	ext{.... (ii)}$
It follows from Eqs. (i) and (ii) that
$\lambda>\lambda^{\prime}>\frac{\lambda}{2}$

Answer

$\lambda^{\prime}<\lambda / 2$

Answer

$\lambda^{\prime}>\lambda / 2$

Answer

$\lambda>\lambda^{\prime}>\frac{\lambda}{2}$

Answer

$\lambda^{\prime}=\frac{\lambda}{2}$

Dual Nature of Matter and Radiation

Light of wavelength \lambda strikes a photoelectric surface and electrons are ejected with kinetic energy K. If K is to be increased to exactly twice its original value, the wavelength must be changed to \lambda^{\prime}, such that

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