AP EAPCET 2024 - 21th May Evening Shift | Dual Nature of Radiation Question 1 | Physics

The maximum wavelength of light which causes photoelectric emission from photosensitive metal surface is $\lambda_0$. Two light beams of wavelengths $\frac{\lambda_0}{3}$ and $\frac{\lambda_0}{9}$ incident on the metal surface. The ratio of the maximum velocities of the emitted photoelectrons is

$3: 4$

$1: 3$

$1: 2$

$2: 3$

AP EAPCET 2024 - 21th May Morning Shift

MCQ (Single Correct Answer)

-0

A blue lamp emits light of mean wavelength $4500$ Å. The lamp is rated at 150 W and $8 \%$ efficiency. Then, the number of photons are emitted by the lamp per second

$27.17 \times 10^{18}$

$17.17 \times 10^{18}$

$27.17 \times 10^{15}$

$54 \times 10^{16}$

AP EAPCET 2024 - 20th May Evening Shift

MCQ (Single Correct Answer)

-0

If the kinetic energy of a particle in motion is decreased by $36 \%$, the increase in de-Broglie wavelength of the particle is

$18 \%$

$25 \%$

$20 \%$

$32 \%$

AP EAPCET 2024 - 20th May Morning Shift

MCQ (Single Correct Answer)

-0

An electron of mass $m$ with initial velocity $\mathbf{v}=v_0 \hat{\mathbf{i}}\left(v_0>0\right)$ enters in an electric field $\mathbf{E}=-E_0 \hat{\mathbf{i}}$ ( $E_0$ is constant $>0$ ) at $t=0$. If $\lambda$ is its de-Broglie wavelength initially, then the de-Broglie wavelength after time $t$ is

$\frac{\lambda}{1+\frac{e E_d}{m v_0}}$

$\frac{\lambda}{\left(1-\frac{e E_d}{m v_0}\right)^2}$

$\left(1+\frac{e E_\alpha}{m v_0}\right) \lambda$

$\left(1+\frac{e E_d}{m v_0}\right)^2 \lambda$

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