Question 383: 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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